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, $13 \times \frac{2}{13}=2$
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
$\therefore K=3$
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:
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$
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:
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$
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$
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
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.
Explanation:
We have
$\cos \alpha = {{2\sqrt 2 } \over 3}$
$\sin \alpha = {1 \over 3}$
$\tan \alpha = {{2\sqrt 2 } \over R}$
$ \Rightarrow R = {{2\sqrt 2 } \over {\tan \alpha }} = 8$ units.
