Circle
Explanation:
$P\left(x_1 y_1\right)$ and point $Q\left(x_2, y_2\right)$
Mid point of $\mathrm{PQ} M=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
Substitute M into $x-y+1=0$
$ x_1+x_2-y_1-y_2+2=0 . .(i) $
Slope of PQ is Perpendicular to slope of bisector line So, slope of $P Q=-1$
$ y_2=x_1-x_2+y_1 \ldots . .(i i) $
$Q\left(x_2, y_2\right)$ lie on $5 x+y+2=0$
So, $5 x_2+y_2+2=0 \ldots \ldots$. (iii)
Substitute (iii) in (i)
$ x_2=\frac{-x_1-y_1-2}{4} \ldots .(i v) $
Substitute (iii) in (ii)
$ x_2=y_1-1 \ldots . .(v) $
From (iv) and (v)
$ x_1=2-5 y $
$\left(x_1, y_1\right)$ lie on circle
$ x_1^2+y_1^2=4 $
Pt $x_1=2-5 y_1$
$ y_1=0,-\frac{10}{13} $
So, $x_1=2, \frac{-24}{13}$
So, $2+\left(-\frac{24}{13}\right)=\frac{2}{13}$
So, $13 \times \frac{2}{13}=2$
Consider the circle C : $x^2+y^2-6 x-8 y-11=0$. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle $x^2+y^2-\alpha x-\beta y-\gamma=0$, then $\alpha+\beta+2 \gamma$ is equal to $\_\_\_\_$ .
Explanation:
Assumptions :
$\boldsymbol{F}(h, \boldsymbol{k})$ is the foot of the ⟂ drop from origin on the variable chord $\boldsymbol{A} \boldsymbol{B}$
The end-points of the chord $\boldsymbol{A B}$ are
$ \begin{array}{|l|l|} \hline \boldsymbol{A} & \boldsymbol{B} \\ \hline\left(\boldsymbol{x}_1, \boldsymbol{y}_1\right) & \left(\boldsymbol{x}_2, \boldsymbol{y}_2\right) \\ \hline \end{array} $
$ \begin{array}{|l|l|l|l|l|} \hline \text { Circle } & \text { Equation } & \text { Transform } & \text { Centre } & \text { Radius } \\ \hline \boldsymbol{C} & \boldsymbol{x}^2+\boldsymbol{y}^2-6 \boldsymbol{x}-8 \boldsymbol{y}-11=0 & (\boldsymbol{x}-3)^2+(\boldsymbol{y}-4)^2=36 & (3,4) & 6 \\ \hline \end{array} $
We will now form the line equations of the ⟂ drop $\boldsymbol{O F}$ and the variable chord $\boldsymbol{A B}$ :
$ l_{O F}: y=\left(\frac{k}{h}\right) x \Rightarrow l_{A B}: y=-\left(\frac{h}{k}\right) x+c ....(i) $
The foot of the ⟂ on line $\boldsymbol{l}_{\boldsymbol{A} \boldsymbol{B}}$ defined in equation (i) is $\boldsymbol{F}(\boldsymbol{h}, \boldsymbol{k})$ :
$ k=-\left(\frac{\boldsymbol{h}}{\boldsymbol{k}}\right) \cdot(\boldsymbol{h})+c \Rightarrow c=\frac{\boldsymbol{h}^2+k^2}{k} \Rightarrow l_{A B}: \boldsymbol{h} \boldsymbol{x}+\boldsymbol{k} \boldsymbol{y}=h^2+k^2 \ldots \ldots (ii) $
Equation (ii) which gives the equation of the variable chord $\boldsymbol{A} \boldsymbol{B}$ cuts the circle at $\boldsymbol{A}\left(\boldsymbol{x}_1, \boldsymbol{y}_1\right)$ and $\boldsymbol{B}\left(\boldsymbol{x}_2, \boldsymbol{y}_2\right)$ is substituted in the equation of the circle $\boldsymbol{C}$ from chart 1 to form 2 quadratics in $\boldsymbol{x}$ and $\boldsymbol{y}$ respectively :
Quadratic in $x$ :
$\boldsymbol{x}^2+\boldsymbol{y}^2-6 \boldsymbol{x}-8 \boldsymbol{y}-11=0 \Rightarrow \boldsymbol{k}^2 x^2+\boldsymbol{k}^2 \boldsymbol{y}^2-6 \boldsymbol{k}^2 \boldsymbol{x}-8 \boldsymbol{k}^2 \boldsymbol{y}-11 \boldsymbol{k}^2=0$
$\boldsymbol{k}^2 \boldsymbol{x}^2+\left(\boldsymbol{h}^2+\boldsymbol{k}^2-\boldsymbol{h} \boldsymbol{x}\right)^2-6 \boldsymbol{k}^2 \boldsymbol{x}-8 \boldsymbol{k}\left(\boldsymbol{h}^2+\boldsymbol{k}^2-\boldsymbol{h} \boldsymbol{x}\right)-11 \boldsymbol{k}^2=0$
$\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right) \boldsymbol{x}^2-2 \boldsymbol{x}\left\{\boldsymbol{h}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)+3 \boldsymbol{k}^2-4 \boldsymbol{h} \boldsymbol{k}\right\}+\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-8 \boldsymbol{k}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{k}^2=0 \leftarrow$ Quadratic in $\boldsymbol{x}$
$\boldsymbol{x}_1 \boldsymbol{x}_2=\frac{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-8 \boldsymbol{k}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{k}^2}{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)} \ldots$. (iii) where $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$ are the roots of the quadratic
Quadratic in $y$ :
$\boldsymbol{x}^2+\boldsymbol{y}^2-6 \boldsymbol{x}-8 \boldsymbol{y}-11=0 \Rightarrow \boldsymbol{h}^2 \boldsymbol{x}^2+\boldsymbol{h}^2 \boldsymbol{y}^2-6 \boldsymbol{h}^2 \boldsymbol{x}-8 \boldsymbol{h}^2 \boldsymbol{y}-11 \boldsymbol{h}^2=0$
$\left(\boldsymbol{h}^2+\boldsymbol{k}^2-\boldsymbol{k} \boldsymbol{y}\right)^2+\boldsymbol{h}^2 \boldsymbol{y}^2-6 \boldsymbol{h}\left(\boldsymbol{h}^2+\boldsymbol{k}^2-\boldsymbol{k} \boldsymbol{y}\right)-8 \boldsymbol{h}^2 \boldsymbol{y}-11 \boldsymbol{h}^2=0$
$\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right) \boldsymbol{y}^2-2 \boldsymbol{y}\left\{\boldsymbol{k}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)+4 \boldsymbol{h}^2-3 \boldsymbol{h} \boldsymbol{k}\right\}+\left(\boldsymbol{h}^2+k^2\right)^2-6 \boldsymbol{h}\left(\boldsymbol{h}^2+k^2\right)-11 \boldsymbol{h}^2=0$ Quadratic in $\boldsymbol{y}$
$\boldsymbol{y}_1 \boldsymbol{y}_2=\frac{\left(h^2+k^2\right)^2-6 h\left(h^2+k^2\right)-11 h^2}{h^2+k^2} \ldots .(\boldsymbol{i v})$ where $\boldsymbol{y}_1$ and $\boldsymbol{y}_2$ are the roots of the quadratic
The variable chord $\boldsymbol{A B}$ with end-points $\boldsymbol{A}\left(\boldsymbol{x}_1, \boldsymbol{y}_1\right)$ and $\boldsymbol{B}\left(\boldsymbol{x}_2, \boldsymbol{y}_2\right)$ subtend a right $\measuredangle$ at origin.
We now infer from equations (iii) and (iv) :
$\left(\frac{y_1}{x_1}\right) \cdot\left(\frac{y_2}{x_2}\right)=-1$
$\Rightarrow \frac{\left\{\frac{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-6 \boldsymbol{h}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{h}^2}{\boldsymbol{h}^2+\boldsymbol{k}^2}\right\}}{\left\{\frac{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-8 \boldsymbol{k}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{k}^2}{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)}\right\}}=-1$
$\Rightarrow \frac{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-6 \boldsymbol{h}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{h}^2}{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-8 \boldsymbol{k}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{k}^2}=-1$
$\Rightarrow $ $ \left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-6 \boldsymbol{h}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{h}^2=-\left\{\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-8 \boldsymbol{k}\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)-11 \boldsymbol{k}^2\right\} $
$\Rightarrow $ $2\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)^2-\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)(6 \boldsymbol{h}+8 \boldsymbol{k})-11\left(\boldsymbol{h}^2+\boldsymbol{k}^2\right)=0$
$\Rightarrow $ $2\left(h^2+k^2\right) \cdot\left(h^2+k^2-3 h-4 k-\frac{11}{2}\right)=0$
$\Rightarrow $ $\boldsymbol{h}^2+\boldsymbol{k}^2-3 \boldsymbol{h}-4 \boldsymbol{k}-\frac{11}{2}=0 \leftarrow \boldsymbol{h}^2+\boldsymbol{k}^2 \neq 0[\because \boldsymbol{F}(\boldsymbol{h}, \boldsymbol{k})$ is not the origin $]$
$\Rightarrow $ $x^2+y^2-3 x-4 y-\frac{11}{2}=0 \ldots . .(\boldsymbol{v}) \leftarrow$ Locus of $F(h, k)$
The circle $C: \boldsymbol{x}^2+\boldsymbol{y}^2-\boldsymbol{\alpha} \boldsymbol{x}-\boldsymbol{\beta} \boldsymbol{y}-\boldsymbol{\lambda}=0$ be the locus of $\boldsymbol{F}(\boldsymbol{h}, \boldsymbol{k})$ which must be equivalent to equation $(\boldsymbol{v})$ :
$\therefore $ $\boldsymbol{x}^2+\boldsymbol{y}^2-\boldsymbol{\alpha} \boldsymbol{x}-\boldsymbol{\beta} \boldsymbol{y}-\boldsymbol{\lambda} \equiv \boldsymbol{x}^2+\boldsymbol{y}^2-3 \boldsymbol{x}-4 \boldsymbol{y}-\frac{11}{2}$
$ \Rightarrow \begin{array}{|l|l|l|l|l|} \hline \boldsymbol{\alpha} & \boldsymbol{\beta} & \boldsymbol{\lambda} & 2 \boldsymbol{\lambda} & \boldsymbol{\alpha}+\boldsymbol{\beta}+2 \boldsymbol{\lambda} \\ \hline 3 & 4 & \frac{11}{2} & 11 & 18 \\ \hline \end{array} $
Let the line $x-y=4$ intersect the circle $\mathrm{C}:(x-4)^2+(y+3)^2=9$ at the points Q and R . If $\mathrm{P}(\alpha, \beta)$ is a point on C such that $\mathrm{PQ}=\mathrm{PR}$, then $(6 \alpha+8 \beta)^2$ is equal to $\_\_\_\_$ .
Explanation:
Since $PQ = PR$, the point $P$ must lie on the perpendicular bisector of the chord $QR$.
Now, the points $Q$ and $R$ are the intersection points of the line
$ x-y=4 $
with the circle
$ (x-4)^2+(y+3)^2=9. $
So, $QR$ is a chord of the circle lying on the line $x-y=4$.
For a circle, the perpendicular bisector of any chord passes through the centre.
The centre of the circle is
$ (4,-3). $
The slope of the line $x-y=4$ is $1$, so the slope of the perpendicular bisector is $-1$.
Hence the perpendicular bisector of chord $QR$ passing through $(4,-3)$ is
$ y+3=-1(x-4) $
$ y+3=-x+4 $
$ x+y=1. $
Therefore, point $P(\alpha,\beta)$ lies on both:
- the circle:
$ (\alpha-4)^2+(\beta+3)^2=9 $
- the line:
$ \alpha+\beta=1. $
Now we need $(6\alpha+8\beta)^2$.
Using $\alpha+\beta=1$, let
$ \beta=1-\alpha. $
Substitute into the circle:
$ (\alpha-4)^2+((1-\alpha)+3)^2=9 $
$ (\alpha-4)^2+(4-\alpha)^2=9 $
Since $(4-\alpha)^2=(\alpha-4)^2$,
$ 2(\alpha-4)^2=9 $
$ (\alpha-4)^2=\frac{9}{2}. $
Now,
$ 6\alpha+8\beta=6\alpha+8(1-\alpha)=8-2\alpha. $
So,
$ (6\alpha+8\beta)^2=(8-2\alpha)^2=4(\alpha-4)^2. $
Using $(\alpha-4)^2=\frac{9}{2}$,
$ (6\alpha+8\beta)^2=4\cdot \frac{9}{2}=18. $
Hence, the required value is
$ \boxed{18}. $
Let the centre of the circle $x^2+y^2+2 \mathrm{~g} x+2 f y+25=0$ be in the first quadrant and lie on the line $2 x-y=4$. Let the area of an equilateral triangle inscribed in the circle be $27 \sqrt{3}$. Then the square of the length of the chord of the circle on the line $x=1$ is $\_\_\_\_$ .
Explanation:
For the circle
$ x^2+y^2+2gx+2fy+25=0, $
the centre is
$ (-g,-f) $
and radius is
$ r=\sqrt{g^2+f^2-25}. $
We are given that the centre lies in the first quadrant, so
$ (-g,-f) $
is in the first quadrant.
Also, the centre lies on the line
$ 2x-y=4. $
So if the centre is $(-g,-f)$, then
$ 2(-g)-(-f)=4 $
that is,
$ -2g+f=4. $
Now use the information about the equilateral triangle inscribed in the circle.
For an equilateral triangle of side $a$,
$ \text{Area}=\frac{\sqrt{3}}{4}a^2. $
Given area is
$ 27\sqrt{3}. $
So,
$ \frac{\sqrt{3}}{4}a^2=27\sqrt{3} $
which gives
$ a^2=108 \quad \Rightarrow \quad a=6\sqrt{3}. $
For an equilateral triangle inscribed in a circle, the circumradius is
$ R=\frac{a}{\sqrt{3}}. $
Hence,
$ r=\frac{6\sqrt{3}}{\sqrt{3}}=6. $
Therefore,
$ r^2=36. $
So,
$ g^2+f^2-25=36 $
which gives
$ g^2+f^2=61. $
Now solve the system
$ -2g+f=4 $
and
$ g^2+f^2=61. $
From the first equation,
$ f=4+2g. $
Substitute into the second:
$ g^2+(4+2g)^2=61 $
$ g^2+16+16g+4g^2=61 $
$ 5g^2+16g-45=0. $
Solve:
$ 5g^2+16g-45=0. $
Discriminant:
$ \Delta=16^2-4\cdot 5\cdot(-45)=256+900=1156=34^2. $
Thus,
$ g=\frac{-16\pm 34}{10}. $
So,
$ g=\frac{18}{10}=\frac95 \quad \text{or} \quad g=\frac{-50}{10}=-5. $
Now centre is $(-g,-f)$ and must be in the first quadrant.
If $g=\frac95$, then
$ f=4+2\cdot \frac95=\frac{38}{5}, $
so centre is
$ \left(-\frac95,-\frac{38}{5}\right), $
not in first quadrant.
Hence reject.
If $g=-5$, then
$ f=4+2(-5)=-6. $
So centre is
$ (5,6), $
which is in the first quadrant.
Thus the circle is
$ x^2+y^2-10x-12y+25=0, $
with centre $(5,6)$ and radius $6$.
Now we need the chord cut by the line
$ x=1. $
Distance of the line $x=1$ from the centre $(5,6)$ is
$ |5-1|=4. $
If a chord is at distance $d$ from the centre in a circle of radius $r$, then chord length is
$ 2\sqrt{r^2-d^2}. $
So here chord length is
$ 2\sqrt{6^2-4^2}=2\sqrt{36-16}=2\sqrt{20}=4\sqrt{5}. $
Therefore, the square of the chord length is
$ (4\sqrt{5})^2=80. $
Hence, the required answer is
$ \boxed{80}. $
Let a circle C have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of C on the line $x + y = 1$ is $\sqrt{14}$, then the square of the radius of C is ________.
Explanation:
$\begin{aligned} & r^2=\left(\frac{\sqrt{14}}{2}\right)^2+\left(r-\frac{1}{\sqrt{2}}\right)^2 \\ & r^2=\frac{7}{2}+r^2-\sqrt{2} r+\frac{1}{2} \\ & \sqrt{2} r=4 \\ & r=2 \sqrt{2} \\ & r^2=8\end{aligned}$
Let $C$ be the circle $x^2+(y-1)^2=2, E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on x -axis and y -axis respectively. Let the straight line $x+y=3$ touch the curves $C, E_1$ and $E_2$ at $P\left(x_1, y_1\right), Q\left(x_2, y_2\right)$ and $R\left(x_3, y_3\right)$ respectively. Given that $P$ is the mid point of the line segment $Q R$ and $P Q=\frac{2 \sqrt{2}}{3}$, the value of $9\left(x_1 y_1+x_2 y_2+x_3 y_3\right)$ is equal to _______.
Explanation:
Solving the line $x+y=3$, and the circle $x^2+$ $(y-1)^2=2$
Substitute $y=3-x$ :
$\begin{aligned} & x^2+(3-x-1)^2=2 \\ & \Rightarrow x^2-2 x+1=0 \\ & \Rightarrow x=1 \Rightarrow y=2 \end{aligned}$
So, $P=\left(x_1, y_1\right)=(1,2) \Rightarrow x_1 y_1=1 \cdot 2=2$
Use midpoint condition
Let $Q=\left(x_2, y_2\right), R=\left(x_3, y_3\right)$.
Since $P$ is the midpoint of QR:
$x_2+x_3=2 x_1=2, y_2+y_3=2 y_1=4
$So, we can write: $x_3=2-x_2, y_3=4-y_2$
Given,
$P Q=\frac{2 \sqrt{2}}{3} \Rightarrow P Q^2=\left(x_2-1\right)^2+\left(y_2-2\right)^2=\frac{8}{9}$
Let's denote: $x_2=a, y_2=b, x_3=2-a, y_3=4-b$
$\begin{aligned} & (a-1)^2+(b-2)^2=\frac{8}{9} \\ & \Rightarrow a^2-2 a+1+b^2-4 b+4=\frac{8}{9} \\ & \Rightarrow a^2+b^2-2 a-4 b+5=\frac{8}{9} \\ & \Rightarrow 9 a^2+9 b^2-18 a-36 b+37=0 \end{aligned}$
Hence, $a=\frac{5}{3}, b=\frac{4}{3}$
$\begin{aligned} & x_1 y_1+x_2 y_2+x_3 y_3=2+a b+(2-a)(4-b) \\ & 9\left(x_1 y_1+x_2 y_2+x_3 y_3\right)=9(10+2 a b-2 b-4 a) \\ & =90+18 a b-18 b-36 a=46 \end{aligned}$
The absolute difference between the squares of the radii of the two circles passing through the point $(-9,4)$ and touching the lines $x+y=3$ and $x-y=3$, is equal to ________ .
Explanation:
$\because x+y=3$ and $x-y=3$ are tangents
$\therefore \quad$ Both circle centre will lie on $x$-axis
$\therefore(x-a)^2+y^2=r^2$
Hence centre is $C(\alpha, 0)$
$\begin{aligned} &r=\sqrt{(\alpha+9)^2+16}\quad\text{.... (1)}\\ &\text { Also }\left|\frac{\alpha-3}{\sqrt{2}}\right|=r \quad\text{.... (2)}\\ &\begin{aligned} & \sqrt{(\alpha+9)^2+16}=\left|\frac{\alpha-3}{\sqrt{2}}\right| \\ & \Rightarrow \quad \alpha=-5 \text { or }-37 \\ & \mathrm{r}=\left|\frac{-5-3}{\sqrt{2}}\right| \text { or }\left|\frac{-37-3}{\sqrt{2}}\right| \\ & =4 \sqrt{2} \text { or } 20 \sqrt{2} \\ & \left|\mathrm{r}_1^2-\mathrm{r}_2^2\right|=|32-800|=768 \end{aligned} \end{aligned}$
Let the circle $C$ touch the line $x-y+1=0$, have the centre on the positive $x$-axis, and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3 x+2 y=1$. Let H be the hyperbola $\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$, whose one of the foci is the centre of $C$ and the length of the transverse axis is the diameter of $C$. Then $2 \alpha^2+3 \beta^2$ is equal to ________.
Explanation:
$\begin{aligned} &x-y+1=0\\ &\mathrm{p}=\mathrm{r}\\ &\left|\frac{\alpha-0+1}{\sqrt{2}}\right|=r \Rightarrow(\alpha+1)^2=2 r^2\quad\text{.... (1)} \end{aligned}$
$\begin{aligned} & \text { now }\left(\frac{-3 \alpha+0-1}{\sqrt{9+4}}\right)^2+\left(\frac{2}{\sqrt{13}}\right)^2=\mathrm{r}^2 \\ & \Rightarrow(3 \alpha+1)^2+4=13 \mathrm{r}^2 \ldots \ldots .(2) \\ & \text { (1) & }(2) \Rightarrow(3 \alpha+1)^2+4=13 \frac{(\alpha+1)^2}{2} \\ & \quad \Rightarrow 18 \alpha^2+12 \alpha+2+8=13 \alpha^2+26 \alpha+13 \\ & \Rightarrow 5 \alpha^2-14 \alpha-3=0 \\ & \Rightarrow 5 \alpha^2-15 \alpha+\alpha-3=0 \\ & \Rightarrow 5 \alpha^2-15 \alpha+\alpha-3=0 \\ & \Rightarrow \alpha=\frac{-1}{5}, 3 \end{aligned}$
$\begin{aligned} &\therefore \quad r=2 \sqrt{2}\\ &\text { How } \alpha \mathrm{e}=3 \text { and } 2 \alpha=4 \sqrt{2}\\ &\begin{aligned} & \alpha^2 \mathrm{e}^2=9 \Rightarrow \alpha=2 \sqrt{2} \Rightarrow \alpha^2=8 \\ & \alpha^2\left(1+\frac{\beta^2}{\alpha^2}\right)=9 \\ & \alpha^2+\beta^2=9 \\ & \therefore \beta^2=1 \\ & \therefore 2 \alpha^2+3 \beta^2=2(8)+3(1)=19 \end{aligned} \end{aligned}$
Let the centre of a circle, passing through the points $(0,0),(1,0)$ and touching the circle $x^2+y^2=9$, be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k), 4\left(h^2+k^2\right)$ is equal to __________.
Explanation:
Circle will touch internally
$\begin{aligned} & C_1 C_2=\left|r_1-r_2\right| \\ & =\sqrt{h^2+k^2}=3-\sqrt{h^2+k^2} \\ & \Rightarrow 2 \sqrt{h^2+k^2}=3 \\ & \Rightarrow h^2+k^2=\frac{9}{4} \\ & \therefore 4\left(h^2+k^2\right)=9 \end{aligned}$
Consider two circles $C_1: x^2+y^2=25$ and $C_2:(x-\alpha)^2+y^2=16$, where $\alpha \in(5,9)$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $C_1$ and $C_2$ be $\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right)$. If the length of common chord of $C_1$ and $C_2$ is $\beta$, then the value of $(\alpha \beta)^2$ equals _______.
Explanation:
$\begin{gathered} C_1: x^2+y^2=25, C_2:(x-\alpha)^2+y^2=16 \\ 5<\alpha<9 \end{gathered}$
$\begin{aligned} & \theta=\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right) \\ & \sin \theta=\frac{\sqrt{63}}{8} \end{aligned}$
Area of $\triangle \mathrm{OAP}=\frac{1}{2} \times \alpha\left(\frac{\beta}{2}\right)=\frac{1}{2} \times 5 \times 4 \sin \theta$
$\begin{aligned} \Rightarrow \quad & \alpha \beta=40 \times \frac{\sqrt{63}}{8} \\ & \alpha \beta=5 \times \sqrt{63} \\ & (\alpha \beta)^2=25 \times 63=1575 \end{aligned}$
Equations of two diameters of a circle are $2 x-3 y=5$ and $3 x-4 y=7$. The line joining the points $\left(-\frac{22}{7},-4\right)$ and $\left(-\frac{1}{7}, 3\right)$ intersects the circle at only one point $P(\alpha, \beta)$. Then, $17 \beta-\alpha$ is equal to _________.
Explanation:
Centre of circle is $(1,-1)$
Equation of $A B$ is $7 x-3 y+10=0 \ldots$ (i)
Equation of $\mathrm{CP}$ is $3 x+7 y+4=0 \ldots$ (ii)
Solving (i) and (ii)
$\alpha=\frac{-41}{29}, \beta=\frac{1}{29} \quad \therefore 17 \beta-\alpha=2$
Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$, where $\alpha, \beta>0$. If the circle touches the line $y+x=0$ at the point $P$, whose distance from the origin is $4 \sqrt{2}$, then $(\alpha+\beta)^2$ is equal to __________.
Explanation:
$\begin{aligned} & S:(x-\alpha)^2+(y-\beta)^2=50 \\ & C P=r \\ & \left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} \\ & \Rightarrow(\alpha+\beta)^2=100 \end{aligned}$
Two circles in the first quadrant of radii $r_{1}$ and $r_{2}$ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $x+y=2$. Then $r_{1}^{2}+r_{2}^{2}-r_{1} r_{2}$ is equal to ___________.
Explanation:
Where $\mathrm{d}=$ perpendicular distance of centre from line $x+y=2$
$ \begin{aligned} & \Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2 \\\\ & \Rightarrow a^2-\frac{(2 a-2)^2}{2}=1 \Rightarrow 2 a^2-4 a^2+8 a-4=2 \\\\ & \Rightarrow 2 a^2-8 a+6=0 \Rightarrow a^2-4 a+3=0 \\\\ & \therefore r_1+r_2=4 \text { and } r_1 r_2=3 \\\\ & \therefore r_1^2+r_2^2-r_1 r_2=\left(r_1+r_2\right)^2-3 r_1 r_2 \\\\ & =16-9=7 \end{aligned} $
Consider a circle $C_{1}: x^{2}+y^{2}-4 x-2 y=\alpha-5$. Let its mirror image in the line $y=2 x+1$ be another circle $C_{2}: 5 x^{2}+5 y^{2}-10 f x-10 g y+36=0$. Let $r$ be the radius of $C_{2}$. Then $\alpha+r$ is equal to _________.
Explanation:
$ \begin{aligned} & C_1: x^2+y^2-4 x-2 y=\alpha-5 \\\\ & C_1:(x-2)^2+(y-1)^5-5=\alpha-5 \\\\ & C_1:(x-2)^2+(y-1)^2=(\sqrt{\alpha})^2 \end{aligned} $
So, centre and radius of $C_1$ are $(2,1)$ and $\sqrt{\alpha}$ respectively
Now, image of $(2,1)$ along the line $y=2 x+1$ is,
$ \frac{x-2}{2}=\frac{y-1}{-1}=\frac{-2(4-1+1)}{2^2+(-1)^2} $
$ \begin{aligned} & \Rightarrow \frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5} \\\\ & \Rightarrow x=\frac{-6}{5} \text { and } y=\frac{13}{5} \end{aligned} $
Now, $\left(\frac{-6}{5}, \frac{13}{5}\right)$ will be the centre of $C_2$
$ \therefore f=\frac{6}{5} \text { and } g=\frac{-13}{5} $
Now, radius of $\mathrm{C}_2=r=\sqrt{f^2+g^2-\frac{36}{5}}$
$ \begin{aligned} & \Rightarrow r=\sqrt{\frac{36}{25}+\frac{169}{25}-\frac{36}{5}}=1 \\\\ & \because r=1 \text { so, } \alpha=1 \\\\ & \therefore \alpha+r=1+1=2 \end{aligned} $
Concept :
Image of a point $\left(x_1, y_1\right)$ w.r.t. $a x+b y+c=0$ is $(x, y)$, then
$ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{\left(a^2+b^2\right)} $
Let the point $(p, p+1)$ lie inside the region $E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^{2}}, 0 \leq x \leq 3\right\}$. If the set of all values of $\mathrm{p}$ is the interval $(a, b)$, then $b^{2}+b-a^{2}$ is equal to ___________.
Explanation:
$ E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\right\} $
Since, point $(p, p+1)$ lie on line $y=x+1$
$\therefore$ Point of intersection of $y=x+1$ and $y=3-x$
i.e., $x+1=3-x$
$\Rightarrow$ $2 x=2 \Rightarrow x=1$
and $y=2$
and point of intersection of $y=x+1$ and
$ y=\sqrt{9-x^2} $
i.e., $(x+1)^2=9-x^2$
$\begin{array}{lc} &\Rightarrow x^2+1+2 x=9-x^2 \\\\ &\Rightarrow 2 x^2+2 x-8=0 \\\\ &\Rightarrow x^2+x-4=0 \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{1+4(1)(4)}}{2} \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{17}}{2}\end{array}$
$\begin{array}{lll}\Rightarrow & x=\frac{-1+\sqrt{17}}{2}, & \text { (Since, } x \in[0,3]) \\\\ \therefore & p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)\end{array}$
$\Rightarrow a=1, b=\frac{-1+\sqrt{17}}{2}$
$\begin{aligned} & \therefore b^2+b-a \\\\ &= \frac{1+17-2 \sqrt{17}}{4}+\frac{(-1+\sqrt{17})}{2}-1 \\\\ &= \frac{18-2 \sqrt{17}-2+2 \sqrt{17}-4}{4} \\\\ &= \frac{12}{4}=3\end{aligned}$
A circle passing through the point $P(\alpha, \beta)$ in the first quadrant touches the two coordinate axes at the points $A$ and $B$. The point $P$ is above the line $A B$. The point $Q$ on the line segment $A B$ is the foot of perpendicular from $P$ on $A B$. If $P Q$ is equal to 11 units, then the value of $\alpha \beta$ is ___________.
Explanation:
Since, (i) passes through $P(\alpha, \beta)$
$\therefore (\alpha-a)^2+(\beta-a)^2=a^2$
$ \begin{array}{lr} &\Rightarrow \alpha^2+\beta^2-2 \alpha a-2 \beta a+a^2=0 .........(i) \end{array} $
Equation of $A B=\frac{x}{a}+\frac{y}{a}=1$
$\Rightarrow x+y=a$ .........(ii)
Let $Q(p, q)$ be the foot of the perpendicular from $P$ to line (iii)
$ \begin{aligned} & \therefore \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-\alpha)}{(1)^2+(1)^2} \\\\ & \Rightarrow \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-a)}{2} \\\\ & \Rightarrow p-\alpha=\frac{-(\alpha+\beta-a)}{2} \\\\ & \text { and } q-\beta=\frac{-(\alpha+\beta-a)}{2} \end{aligned} $
Now,
$ \begin{aligned} & P Q^2 =(p-\alpha)^2+(q-\beta)^2 \\\\ & =\frac{1}{4}(\alpha+\beta-a)^2+\frac{1}{4}(\alpha+\beta-a)^2 \end{aligned} $
$ \Rightarrow (11)^2=\frac{1}{2}(\alpha+\beta-\alpha)^2 $
$ \begin{array}{ll} &\Rightarrow \alpha^2+\beta^2+a^2+2 \alpha \beta-2 \beta a-2 \alpha a=242 \\\\ &\Rightarrow 2 \alpha \beta=242 \text { [Using Eq. (ii)] }\\\\ &\Rightarrow \alpha \beta=121 \end{array} $
Explanation:
$OC \,\bot \,CP$ and $OC \,\bot \, CQ$
$\Rightarrow PCQ$ is a straight line
$OC = \sqrt {{{(\sqrt 2 )}^2} + {{(\sqrt 3 )}^2}} = \sqrt 5 $
Let $CP = CQ = I$
$[OCP] = {1 \over 2} \times OC \times I = {{\sqrt {35} } \over 2}$
$I = \sqrt 7 $
$OP = OQ = \sqrt {{{(OC)}^2} + {I^2}} = \sqrt {5 + 7} = \sqrt {12} $
$a_1^2 + a_2^2 + b_1^2 + b_2^2 = \left( {a_1^2 + b_2^2} \right) + \left( {a_2^2 + b_2^2} \right)$
$O{P^2} + O{Q^2} = 12 + 12 = 24$
A circle with centre (2, 3) and radius 4 intersects the line $x+y=3$ at the points P and Q. If the tangents at P and Q intersect at the point $S(\alpha,\beta)$, then $4\alpha-7\beta$ is equal to ___________.
Explanation:
The line $x + y = 3$ ..... (i)
is polar of $S(\alpha ,\beta )$ w.r.t. circle
${(x - 2)^2} + {(y - 3)^2} = 16$
$ \Rightarrow {x^2} + {y^2} - 4x - 6y - 3 = 0$
Equation of polar is
$\alpha x + \beta y - 2(x + \alpha ) - 3(4 + \beta ) - 3 = 0$
$(\alpha - 2)x + (\alpha - 3)y - (2\alpha + 3\beta + 3) = 0$ ..... (ii)
(i) and (ii) represent the same.
$\therefore$ ${{\alpha - 2} \over 1} = {{\beta - 3} \over 1} = {{2\alpha + 3\beta + 3} \over 3}$
$\alpha - \beta + 1 = 0$
$\alpha - 3\beta - 9 = 0$
$ \Rightarrow \alpha = - 6,\beta = - 5$
$4\alpha - 7\beta = 11$
Points P($-$3, 2), Q(9, 10) and R($\alpha,4$) lie on a circle C and PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line $2x-ky=1$, then k is equal to ____________.
Explanation:
Now, $\frac{10-2}{9+3} \times \frac{10-4}{9-\alpha}=-1$
$\Rightarrow \frac{8}{12} \cdot 6=\alpha-9 \Rightarrow \alpha=13$
$\therefore 0=(5,3)$ So, $ m_{O Q}=\frac{7}{4} $ $ \begin{aligned} & m_{O R}=\frac{1}{8} \end{aligned} $
$\therefore Q: y-10=\frac{-4}{7}(x-9)$
$\Rightarrow 4 x+7 y=106$
Tangent at $R: y-4=-8(x-13)$
$ 8 x+y=108\quad...(ii) $
By (i) and (ii) $S \equiv\left(\frac{25}{2}, 8\right)$, satisfies with the line
$\therefore K=3$
Explanation:
Let say, $Z=(2)(1)^{1 / 8}$
$ \begin{aligned} & \Rightarrow Z^8=2^8 \times 1 \\\\ & \Rightarrow Z^8-2^8=0 \end{aligned} $
$\begin{aligned} & \Rightarrow Z=2,2 \alpha, 2 \alpha^2, 2 \alpha^3, \ldots, 2 \alpha^7 ; \alpha=e^{i \frac{2 \pi}{8}} \\\\ & \Rightarrow Z^8-2^8=(Z-2)(Z-2 \alpha)\left(Z-2 \alpha^2\right)\left(Z-2 \alpha^3\right) \ldots\left(Z-2 \alpha^7\right) \\\\ & \Rightarrow\left|Z^8-2^8\right|=|Z-2||Z-2 \alpha| \ldots .\left|Z-2 \alpha^7\right| \\\\ & \text { But }\left|Z^8+\left(-2^8\right)\right| \leq|Z|^8+2^8\end{aligned}$
$\begin{aligned} \Rightarrow|Z-2||Z-2 \alpha| \ldots\left|Z-2 \alpha^7\right| & \leq|Z|^8+2^8 \\\\ & \leq 2^8+2^8 \\\\ & \leq 2^9\end{aligned}$
$\Rightarrow \operatorname{Max}\left(P A_1 \cdot P A_2 \ldots P A_8\right)=2^9$
Explanation:
Let $M$ and $N$ be midpoints of $P Q$ and $S T$ respectively.
$\Rightarrow M N$ is a radical axis of two circles
$C_1: x^2+y^2=1$ ........(i)
$\begin{aligned} & C_2:(x-4)^2+(y-1)^2=r^2 \\\\ & \Rightarrow x^2+y^2-8 x-2 y+17-r^2=0 .......(ii)\end{aligned}$
From (i) and (ii);
Equation of $M N: 8 x+2 y-18+r^2=0$
$\Rightarrow B$ is on $x$-axis $\Rightarrow B\left(\frac{18-r^2}{8}, 0\right)$
$ \begin{aligned} & A B=\sqrt{5} \\\\ & \sqrt{\left(\frac{18-r^2}{8}-4\right)^2+1}=\sqrt{5} \end{aligned} $
$\Rightarrow$ On solving $r^2=2$
Let $A B$ be a chord of length 12 of the circle $(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$. If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$, then five times the distance of point $P$ from chord $A B$ is equal to __________.
Explanation:
$ O M=\sqrt{\left(\frac{13}{2}\right)^{2}-6^{2}}=\frac{5}{2} $
$ \sin \theta=\frac{12}{13} $
In $\triangle P A O$ :
$ \begin{aligned} &\frac{P O}{O A}=\sec \theta \\\\ &P O=\frac{13}{2} \cdot \frac{13}{5}=\frac{169}{10} \\\\ &\therefore P M=\frac{169}{10}-\frac{5}{2}=\frac{144}{10}=\frac{72}{5} \\\\ &\therefore 5 P M=72 . \end{aligned} $
$\text { Let } S=\left\{(x, y) \in \mathbb{N} \times \mathbb{N}: 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $T=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:(x-7)^{2}+(y-4)^{2} \leq 36\right\}$. Then $n(S \cap T)$ is equal to __________.
Explanation:
represents all the integral points inside
and on the ellipse $\frac{(x-3)^{2}}{16}+\frac{(y-4)^{2}}{9}=1$, in first quadrant.
and $T=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ represents all the points on
and inside the circle $(x-7)^{2}+(y-4)^{2}=36$
$\therefore \quad n(S \cap T)=\{(3,1),(2,2),(3,2),(4,2),(5,2)$, $(2,3), \ldots(6,5)\}$
Total number of points $=27$
Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2 x-6 y+\alpha=0$ in line $y=x+1$ be $c_{2}: 5 x^{2}+5 y^{2}+10 g x+10 f y+38=0$. If $\mathrm{r}$ is the radius of circle $\mathrm{c}_{2}$, then $\alpha+6 \mathrm{r}^{2}$ is equal to ________.
Explanation:
${c_1}:{x^2} + {y^2} - 2x - 6y + \alpha = 0$
Then centre $ = (1,3)$ and radius $(r) = \sqrt {10 - \alpha } $
Image of $(1,3)$ w.r.t. line $x - y + 1 = 0$ is $(2,2)$
${c_2}:5{x^2} + 5{y^2} + 10gx + 10fy + 38 = 0$
or ${x^2} + {y^2} + 2gx + 2fy + {{38} \over 5} = 0$
Then $( - g, - f) = (2,2)$
$\therefore$ $g = f = - 2$ .......... (i)
Radius of ${c_2} = r = \sqrt {4 + 4 - {{38} \over 5}} = \sqrt {10 - \alpha } $
$ \Rightarrow {2 \over 5} = 10 - \alpha $
$\therefore$ $\alpha = {{48} \over 5}$ and $r = \sqrt {{2 \over 5}} $
$\therefore$ $\alpha + 6{r^2} = {{48} \over 5} + {{12} \over 5} = 12$
If the circles ${x^2} + {y^2} + 6x + 8y + 16 = 0$ and ${x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y = k + 6\sqrt 3 + 8\sqrt 6 $, $k > 0$, touch internally at the point $P(\alpha ,\beta )$, then ${\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2}$ is equal to ________________.
Explanation:
The circle ${x^2} + {y^2} + 6x + 8y + 16 = 0$ has centre $( - 3, - 4)$ and radius 3 units.
The circle ${x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y = k + 6\sqrt 3 + 8\sqrt 6 ,\,k > 0$ has centre $\left( {\sqrt 3 - 3,\,\sqrt 6 - 4} \right)$ and radius $\sqrt {k + 34} $
$\because$ These two circles touch internally hence
$\sqrt {3 + 6} = \left| {\sqrt {k + 34} - 3} \right|$
Here, $k = 2$ is only possible ($\because$ $k > 0$)
Equation of common tangent to two circles is $2\sqrt 3 x + 2\sqrt 6 y + 16 + 6\sqrt 3 + 8\sqrt 6 + k = 0$
$\because$ $k = 2$ then equation is
$x + \sqrt 2 y + 3 + 4\sqrt 2 + 3\sqrt 3 = 0$ ...... (i)
$\because$ ($\alpha$, $\beta$) are foot of perpendicular from $( - 3, - 4)$
To line (i) then
${{\alpha + 3} \over 1} = {{\beta + 4} \over {\sqrt 2 }} = {{ - \left( { - 3 - 4\sqrt 2 + 3 + 4\sqrt 2 + 3\sqrt 3 } \right)} \over {1 + 2}}$
$\therefore$ $\alpha + 3 = {{\beta + 4} \over {\sqrt 2 }} = - \sqrt 3 $
$ \Rightarrow {\left( {\alpha + \sqrt 3 } \right)^2} = 9$ and ${\left( {\beta + \sqrt 6 } \right)^2} = 16$
$\therefore$ ${\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2} = 25$
If one of the diameters of the circle ${x^2} + {y^2} - 2\sqrt 2 x - 6\sqrt 2 y + 14 = 0$ is a chord of the circle ${(x - 2\sqrt 2 )^2} + {(y - 2\sqrt 2 )^2} = {r^2}$, then the value of r2 is equal to ____________.
Explanation:
$ \text { Radius }=\sqrt{(\sqrt{2})^{2}+(3 \sqrt{2})^{2}-14}=\sqrt{6} $
$\Rightarrow$ Diameter $=2 \sqrt{6}$
If this diameter is chord to
$ \begin{aligned} &(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} \text { then } \\\\ &\Rightarrow r^{2}=6+\left(\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}\right)^{2} \\\\ &\Rightarrow r^{2}=6+4=10 \\\\ &\Rightarrow r^{2}=10 \end{aligned} $
Let the lines $y + 2x = \sqrt {11} + 7\sqrt 7 $ and $2y + x = 2\sqrt {11} + 6\sqrt 7 $ be normal to a circle $C:{(x - h)^2} + {(y - k)^2} = {r^2}$. If the line $\sqrt {11} y - 3x = {{5\sqrt {77} } \over 3} + 11$ is tangent to the circle C, then the value of ${(5h - 8k)^2} + 5{r^2}$ is equal to __________.
Explanation:
${L_1}:y + 2x = \sqrt {11} + 7\sqrt 7 $
${L_2}:2y + x = 2\sqrt {11} + 6\sqrt 7 $
Point of intersection of these two lines is centre of circle i.e. $\left( {{8 \over 3}\sqrt 7 ,\sqrt {11} + {5 \over 3}\sqrt 7 } \right)$
${ \bot ^r}$ from centre to line $3x - \sqrt {11} y + \left( {{{5\sqrt {77} } \over 3} + 11} \right) = 0$ is radius of circle
$ \Rightarrow r = \left| {{{8\sqrt 7 - 11 - {5 \over 3}\sqrt {77} + {{5\sqrt {77} } \over 3} + 11} \over {\sqrt {20} }}} \right|$
$ = \left| {\root 4 \of {{7 \over 5}} } \right| = \root 4 \of {{7 \over 5}} $ units
So ${(5h - 8K)^2} + 5{r^2}$
$ = {\left( {{{40} \over 3}\sqrt 7 - 8\sqrt {11} - {{40} \over 3}\sqrt 7 } \right)^2} + 5.\,16.\,{7 \over 5}$
$ = 64 \times 11 + 112 = 816$.
Let a circle C of radius 5 lie below the x-axis. The line L1 : 4x + 3y + 2 = 0 passes through the centre P of the circle C and intersects the line L2 = 3x $-$ 4y $-$ 11 = 0 at Q. The line L2 touches C at the point Q. Then the distance of P from the line 5x $-$ 12y + 51 = 0 is ______________.
Explanation:
${L_1}:4x + 3y + 2 = 0$
${L_2}:3x - 4y - 11 = 0$
Since circle C touches the line L2 at Q intersection point Q of L1 and L2, is (1, $-$2)
$\because$ P lies of L1
$\therefore$ $P\left( {x, - {1 \over 3}(2 + 4x)} \right)$
Now, $PQ = 5 \Rightarrow {(x - 1)^2} + {\left( {{{4x + 2} \over 3} - 2} \right)^2} = 25$
$ \Rightarrow {(x - 1)^2}\left[ {1 + {{16} \over 9}} \right] = 25$
$ \Rightarrow {(x - 1)^2} = 9$
$ \Rightarrow x = 4,\, - 2$
$\because$ Circle lies below the x-axis
$\therefore$ y = $-$6
P(4, $-$6)
Now distance of P from 5x $-$ 12y + 51 = 0
$ = \left| {{{20 + 72 + 51} \over {13}}} \right| = {{143} \over {13}} = 11$
A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is 2x $-$ y + 4 = 0, then the area of R is ____________.
Explanation:
As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side
$\therefore$ $a = \sqrt {{{(3 - 1)}^2} + {{(6 - 2)}^2}} = \sqrt {20} $
and $b/2 = {4 \over {\sqrt 5 }} \Rightarrow b = {8 \over {\sqrt 5 }}$
Area $ = ab = 2\sqrt 5 \,.\,{8 \over {\sqrt 5 }} = 16$.
Let the abscissae of the two points P and Q be the roots of $2{x^2} - rx + p = 0$ and the ordinates of P and Q be the roots of ${x^2} - sx - q = 0$. If the equation of the circle described on PQ as diameter is $2({x^2} + {y^2}) - 11x - 14y - 22 = 0$, then $2r + s - 2q + p$ is equal to __________.
Explanation:
Let $P({x_1},{y_1})$ & $Q({x_2},{y_2})$
$\therefore$ Roots of $2{x^2} - rx + p = 0$ are ${x_1},\,{x_2}$
and roots of ${x^2} - sx - q = 0$ are ${y_1},\,{y_2}$.
$\therefore$ Equation of circle $ \equiv (x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$
$ \Rightarrow {x^2} - ({x_1} + {x_2})x + {x_1}{x_2} + {y^2} - ({y_1} + {y_2})y + {y_1}{y_2} = 0$
$ \Rightarrow {x^2} - {r \over 2}x + {p \over 2} + {y^2} + sy - q = 0$
$ \Rightarrow 2{x^2} + 2{y^2} - rx + 2sy + p - 2q = 0$
Compare with $2{x^2} + 2{y^2} - 11x - 14y - 22 = 0$
We get $r = 11,\,s = 7,\,p - 2q = - 22$
$ \Rightarrow 2r + s + p - 2q = 22 + 7 - 22 = 7$
Let a circle C : (x $-$ h)2 + (y $-$ k)2 = r2, k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to ___________.
Explanation:
Here, $O{M^2} = O{P^2} - P{M^2}$
${\left( {{{|1 + r|} \over {\sqrt 2 }}} \right)^2} = {r^2} - 1$
$\therefore$ ${r^2} - 2r - 3 = 0$
$\therefore$ $r = 3$
$\therefore$ Equation of circle is
${(x - 1)^2} + {(y - 3)^2} = {3^2}$
$\therefore$ h = 1, k = 3, r = 3
$\therefore$ $h + k + r = 7$
Explanation:
Here ABC is a right angle triangle. BC is the Hypotenuse of the triangle.
We know, diameter of circumcircle of a right angle triangle is equal to the Hypotenuse of the triangle also midpoint of Hypotenuse is the center of circle.
$\therefore$ $BC$ = Diameter of the circle
Here $B = (0,1)$ and $C(3,0)$
$\therefore$ $BC = \sqrt {{3^2} + {1^2}} $
$ = \sqrt {9 + 1} $
$ = \sqrt {10} $
$\therefore$ Radius of circumcircle $(R) = {{\sqrt {10} } \over 2}$
$\therefore$ Center of circle $(M) = \left( {{{3 + 0} \over 2},\,{{0 + 1} \over 2}} \right) = \left( {{3 \over 2},\,{1 \over 2}} \right)$
Center of circle which touches line AB and AC $ = (r,r)$
Now distance between center of two circles,
$ME = R - r = {{\sqrt {10} } \over 2} - r$
$ \Rightarrow {\left( {r - {3 \over 2}} \right)^2} + {\left( {r - {1 \over 2}} \right)^2} = {\left( {{{\sqrt {10} } \over 2} - r} \right)^2}$
$ \Rightarrow {r^2} - 3r + {9 \over 4} + {r^2} - r + {1 \over 4} = {{10} \over 4} + {r^2} - \sqrt {10} r$
$ \Rightarrow {r^2} - 4r + \sqrt {10} r = 0$
$ \Rightarrow r(r - 4 + \sqrt {10} ) = 0$
$ \Rightarrow r = 0$ or $r = r - \sqrt {10} $
$\therefore$ $r = 4 - \sqrt {10} $ [as $r \ne 0$]
$ = 0.837$
$ \simeq 0.84$
Explanation:
Radius = $\sqrt {1 + 4 - 1} = 2$
$AB = \sqrt {{3^2} + {2^2}} = \sqrt {13} $
In $\Delta$ABP
$A{P^2} = A{B^2} - B{P^2} = 13 - 4 = 9$
AP = 3
AQ = AP = 3
Let $\angle$ABP = $\theta$, $\angle$BAP = 90$-$ $\theta$
In $\Delta$ABP, tan$\theta$ = 3/2
$\sin \theta = {3 \over {\sqrt {13} }}$, $\cos \theta = {2 \over {\sqrt {13} }}$
In $\Delta$ARP,
$\cos (90 - \theta ) = {{AR} \over {AP}} \Rightarrow AR = 3\sin \theta $
In $\Delta$BRP,
$\cos \theta = {{BR} \over {BP}}$
$ \Rightarrow BR = 2\cos \theta = {{Area\,(\Delta APQ)} \over {Area\,(\Delta BPQ)}} = {{{1 \over 2} \times PQ \times AR} \over {{1 \over 2} \times PQ \times BR}}$
$ = {{AR} \over {RB}} = {{3\sin \theta } \over {2\cos \theta }} = {9 \over 4}$
$ \Rightarrow 8\left( {{{Area\,(\Delta APQ)} \over {Area\,(\Delta BPQ)}}} \right) = 18$
circles (x $-$ 1)2 + (y $-$ 1)2 = 1
and (x $-$ 9)2 + (y $-$ 1)2 = 4, without intercepting a chord on either circle, then the sum of all the integral values of $\alpha$ is ___________.
Explanation:
Both centers should lie on either side of the line as well as line can be tangent to circle.
(3 + 4 $-$ $\alpha$) . (27 + 4 $-$ $\alpha$) < 0
(7 $-$ $\alpha$) . (31 $-$ $\alpha$) < 0 $\Rightarrow$ $\alpha$ $\in$ (7, 31) ....... (1)
d1 = distance of (1, 1) from line
d2 = distance of (9, 1) from line
${d_1} \ge {r_1} \Rightarrow {{|7 - \alpha |} \over 5} \ge 1 \Rightarrow \alpha \in ( - \infty ,2] \cup [12,\infty )$ .... (2)
${d_2} \ge {r_2} \Rightarrow {{|31 - \alpha |} \over 5} \ge 2 \Rightarrow \alpha \in ( - \infty ,21] \cup [41,\infty )$ ....(3)
(1) $\cap$ (2) $\cap$ (3) $\Rightarrow$ $\alpha$ $\in$ [12, 21]
Sum of integers = 165
Explanation:
$ \Rightarrow \cos \theta = {3 \over 5},\sin \theta = {4 \over 5}$
Now using parametric form
${{x - 1} \over {\cos \theta }} = {{y - 2} \over {\sin \theta }} = \pm \,5$
(x, y) = (1 + 5cos$\theta$, 2 + 5sin$\theta$)
($\alpha$, $\beta$) = (4, 6)
(x, y) = ($\gamma$, $\delta$) = (1 $-$ 5cos$\theta$, 2 $-$ 5sin$\theta$)
($\gamma$, s) = ($-$2, $-$2)
$\Rightarrow$ |($\alpha$ + $\beta$) ($\gamma$ + $\delta$)| = | 10x $-$ 4 | = 40
Explanation:
Since, $r \in (0,5]$
So, $0 < 2{p^2} - 2p - 19 \le 100$
$ \Rightarrow p \in \left[ {{{1 - \sqrt {239} } \over 2},{{1 - \sqrt {39} } \over 2}} \right) \cup \left( {{{1 + \sqrt {39} } \over 2},{{1 + \sqrt {239} } \over 2}} \right]$
so, number of integral values of p2 is 61.
Explanation:
A(0, 0), B(1, 0), C(0, 1), D(1, 1)
(PA)2 + (PB)2 + (PC)2 + (PD)2 = 18
${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$ = 18
$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$
$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$
$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $
$ \Rightarrow {d^2} = 16$
x2 + y2 $-$ 10x $-$ 10y + 41 = 0
x2 + y2 $-$ 24x $-$ 10y + 160 = 0 is ___________.
Explanation:
Centre (5, 5), r1 = 3
${S_2}:{(x - 12)^2} + {(y - 5)^2} = 9$
Centre (12, 5), r2 = 3
So (P1P2)min = 1
Explanation:
Given
PA = 3PB
PA2 = 9PB2
$ \Rightarrow $ (h $-$ 5)2 + k2 = 9[(h + 5)2 + k2]
$ \Rightarrow $ 8h2 + 8k2 + 100h + 200 = 0
$ \therefore $ Locus
${x^2} + {y^2} + \left( {{{25} \over 2}} \right)x + 25 = 0$
$ \therefore $ $c \equiv \left( {{{ - 25} \over 4},0} \right)$
$ \therefore $ ${r^2} = {\left( {{{ - 25} \over 4}} \right)^2} - 25$
$ = {{625} \over {16}} - 25$
$ = {{225} \over {16}}$
$ \therefore $ $4{r^2} = 4 \times {{225} \over {16}} = {{225} \over 4} = 56.25$
After Round of 4r2 = 56
Explanation:
Explanation:
Let centre O2 (2, 1) of required circle and its radius being r.
Distance between (1, 3) and (2, 1) is $\sqrt 5 $
$ \therefore $ ${\left( {\sqrt 5 } \right)^2} + {(2)^2} = {r^2}$
$ \Rightarrow r = 3$
The radius of the circle C is ___________.
Explanation:
Let equation of circle be
(x $-$ h)2 + y2 = h2 .... (i)
Solving Eq. (i) with y2 = 4 $-$ x, we get
x2 $-$ 2hx + 4 $-$ x = 0
$\Rightarrow$ x2 $-$ x(2h + 1) + 4 = 0 .... (ii)
For touching/tangency, Discriminant (D) = 0
i.e. (2h + 1)2 = 16 $\Rightarrow$ 2h + 1 = $\pm$ 4
$\Rightarrow$ 2h = $\pm$ 4 $-$ 1
$\Rightarrow$ $h = {3 \over 2},h = {{ - 5} \over 2}$ (Rejected) because part of circle lies outside R. So, $h = {3 \over 2}$ = radius of circle (C).
The value of $\alpha$ is ___________.
Explanation:
Let equation of circle be
(x $-$ h)2 + y2 = h2 .... (i)
Solving Eq. (i) with y2 = 4 $-$ x, we get
x2 $-$ 2hx + 4 $-$ x = 0
$\Rightarrow$ x2 $-$ x(2h + 1) + 4 = 0 .... (ii)
For touching/tangency, Discriminant (D) = 0
i.e. (2h + 1)2 = 16 $\Rightarrow$ 2h + 1 = $\pm$ 4
$\Rightarrow$ 2h = $\pm$ 4 $-$ 1
$\Rightarrow$ $h = {3 \over 2},h = {{ - 5} \over 2}$ (Rejected) because part of circle lies outside R. So, $h = {3 \over 2}$ = radius of circle (c).
Putting h = 3/2 in Eq. (ii),
x2 $-$ 4x + 4 = 0 $\Rightarrow$ (x $-$ 2)2 = 0 $\Rightarrow$ x = 2
So, $\alpha$ = 2
x + y = 2 respectively, then the maximum value of $\alpha\beta $ is _____.
Explanation:
$Q( - 3\cos \theta ,\, - 3\sin \theta )$
$\alpha = \left| {{{3\cos \theta + 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$
$\beta = \left| {{{ - 3\cos \theta - 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$
$\alpha \beta = \left| {{{{{\left( {3\cos \theta + 3\sin \theta } \right)}^2} - 4} \over 2}} \right|$
$ = \left| {{{5 + 9\sin 2\theta } \over 2}} \right|$
$\alpha {\beta _{\max }}$$ = {{5 + 9} \over 2} = 7$ (when sin2$\theta $ = 1)
Explanation:
$ \because $ center lies on x + y = 2 and in 1st quadrant center = ($\alpha $, 2 $-$ $\alpha $)
where $\alpha $ > 0 and 2 $-$ $\alpha $ > 0 $ \Rightarrow $ 0 < $\alpha $ < 2
$ \because $ circle touches x = 3 and y = 2
$ \therefore $ ${{\left| {\alpha - 3} \right|} \over 1} = r$
and ${{\left| {2 - (2 - \alpha )} \right|} \over 1} = r$
$ \Rightarrow \,|\alpha |\, = r$
$ \therefore $ $|\alpha - 3|\, = \,|\alpha |$
$ \Rightarrow $ ${\alpha ^2} - 6\alpha + 9 = {\alpha ^2}$
$ \Rightarrow \alpha = {3 \over 2}$
$ \therefore $ $r = {3 \over 2}$
$ \Rightarrow $ 2r = 3 = diameter.
x2 + y2 – 2x – 4y + 4 = 0 at two distinct points is ______.
Explanation:
$ \Rightarrow $ (x – 1)2 + (y – 2)2 = 1
Centre: (1, 2), radius = 1
Line 3x + 4y – k = 0 intersects the circle at two distinct points.
$ \Rightarrow $ distance of centre from the line < radius
$ \Rightarrow $ $\left| {{{3 \times 1 + 4 \times 2 - k} \over {\sqrt {{3^2} + {4^2}} }}} \right| < 1$
$ \Rightarrow $ |11 - k| < 5
$ \Rightarrow $ 6 < k < 5
$ \Rightarrow $ k $ \in $ {7, 8, 9, ……15} since k $ \in $ I
$ \therefore $ Total 9 integral value of k.
x2 – 8y + y2 + 16 – k = 0, (k > 0) touch each other at a point, then the largest value of k is ______.
Explanation:
C1(3, 0) and r1 = 1
C2 : x2 + y2 – 8y + 16 – k = 0
C2(0, 4) and r2 = $\sqrt k $
Two circles touch each other
$ \therefore $ C1C2 = | r1 $ \pm $ r2 |
$ \Rightarrow $ 5 = | 1 $ \pm $ $\sqrt k $ |
$ \therefore $ 1 + $\sqrt k $ = 5 or $\sqrt k $ - 1 = 5
$ \Rightarrow $ k = 16 or k = 36
So largest value of k = 36.
Explanation:
(x2 + y2 $-$ r2) + $\lambda $(2x + 4y $-$ 5) = 0 ......(i)
Since, the circle (i) passes through the centre of circle
x2 + y2 = r2,
So, $-$ r2 $-$ 5$\lambda $ = 0
or 5$\lambda $ + r2 = 0 ....(ii)
and the centre of circle (i) lies on the line x + 2y = 4, so centre ($-$ $\lambda $, $-$ 2$\lambda $) satisfy the line x + 2y = 4.
Therefore, $-$$\lambda $ $-$4$\lambda $ = 4
$ \Rightarrow $ $-$5$\lambda $ = 4
$ \Rightarrow $ r2 = 4 {from Eq. (ii)}
$ \Rightarrow $ r = 2
Explanation:
From the figure,
$AC = {2 \over {\sin \theta }}$ ....(i)
$ \because $ $\sin \theta = {1 \over {CB}}$ (from $\Delta $CPB) .... (ii)
and $\sin \theta = {2 \over {AC}} = {2 \over {CB + AB}}$ (from $\Delta $CQA) ....(iii)
$ \because $ AB = AM + MB = 2AM [$ \because $ AM = MB]
$ = 2{{\left| {(8 \times 2) - (6 \times 3) - 23} \right|} \over {\sqrt {64 + 36} }}$
$ = {{2 \times 25} \over {10}} = 5.00$
From Eqs. (ii) and (iii), we get
$\sin \theta = {1 \over {CB}} = {2 \over {CB + AB}}$
$ \Rightarrow {1 \over {CB}} = {2 \over {CB + 5}}$ [$ \because $ AB = 5]
$ \Rightarrow CB + 5 = 2CB \Rightarrow CB = 5 = {1 \over {\sin \theta }}$
From the Eq. (i), we get
$AC = {2 \over {\sin \theta }} = 2 \times 5 = 10.00$
If $S = \left\{ {\left( {2,\,{3 \over 4}} \right),\,\left( {{5 \over 2},\,{3 \over 4}} \right),\,\left( {{1 \over 4} - \,{1 \over 4}} \right),\,\left( {{1 \over 8},\,{1 \over 4}} \right)} \right\}$ then the number of points (s) in S lying inside the smaller part is
Explanation:
$L:2x - 3y - 1$
$S:{x^2} + {y^2} - 6$
If ${L_1} > 0$ and ${S_1} < 0$
The point lies in the smaller part. Therefore, $\left( {2,{3 \over 4}} \right)$ and $\left( {{1 \over 4}, - {1 \over 4}} \right)$ lie inside.
