Circle
If the tangents at the points $\mathrm{P}$ and $\mathrm{Q}$ on the circle $x^{2}+y^{2}-2 x+y=5$ meet at the point $R\left(\frac{9}{4}, 2\right)$, then the area of the triangle $\mathrm{PQR}$ is :
Let a circle $C_{1}$ be obtained on rolling the circle $x^{2}+y^{2}-4 x-6 y+11=0$ upwards 4 units on the tangent $\mathrm{T}$ to it at the point $(3,2)$. Let $C_{2}$ be the image of $C_{1}$ in $\mathrm{T}$. Let $A$ and $B$ be the centers of circles $C_{1}$ and $C_{2}$ respectively, and $M$ and $N$ be respectively the feet of perpendiculars drawn from $A$ and $B$ on the $x$-axis. Then the area of the trapezium AMNB is :
Let $y=x+2,4y=3x+6$ and $3y=4x+1$ be three tangent lines to the circle $(x-h)^2+(y-k)^2=r^2$. Then $h+k$ is equal to :
Let the tangents at the points $A(4,-11)$ and $B(8,-5)$ on the circle $x^{2}+y^{2}-3 x+10 y-15=0$, intersect at the point $C$. Then the radius of the circle, whose centre is $C$ and the line joining $A$ and $B$ is its tangent, is equal to :
The points of intersection of the line $ax + by = 0,(a \ne b)$ and the circle ${x^2} + {y^2} - 2x = 0$ are $A(\alpha ,0)$ and $B(1,\beta )$. The image of the circle with AB as a diameter in the line $x + y + 2 = 0$ is :
The locus of the mid points of the chords of the circle ${C_1}:{(x - 4)^2} + {(y - 5)^2} = 4$ which subtend an angle ${\theta _i}$ at the centre of the circle $C_1$, is a circle of radius $r_i$. If ${\theta _1} = {\pi \over 3},{\theta _3} = {{2\pi } \over 3}$ and $r_1^2 = r_2^2 + r_3^2$, then ${\theta _2}$ is equal to :
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 the tangents at two points $\mathrm{A}$ and $\mathrm{B}$ on the circle $x^{2}+\mathrm{y}^{2}-4 x+3=0$ meet at origin $\mathrm{O}(0,0)$. Then the area of the triangle $\mathrm{OAB}$ is :
For $\mathrm{t} \in(0,2 \pi)$, if $\mathrm{ABC}$ is an equilateral triangle with vertices $\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t)$ and $C(a, b)$ such that its orthocentre lies on a circle with centre $\left(1, \frac{1}{3}\right)$, then $\left(a^{2}-b^{2}\right)$ is equal to :
Let $C$ be the centre of the circle $x^{2}+y^{2}-x+2 y=\frac{11}{4}$ and $P$ be a point on the circle. A line passes through the point $\mathrm{C}$, makes an angle of $\frac{\pi}{4}$ with the line $\mathrm{CP}$ and intersects the circle at the points $Q$ and $R$. Then the area of the triangle $P Q R$ (in unit $^{2}$ ) is :
A circle $C_{1}$ passes through the origin $\mathrm{O}$ and has diameter 4 on the positive $x$-axis. The line $y=2 x$ gives a chord $\mathrm{OA}$ of circle $\mathrm{C}_{1}$. Let $\mathrm{C}_{2}$ be the circle with $\mathrm{OA}$ as a diameter. If the tangent to $\mathrm{C}_{2}$ at the point $\mathrm{A}$ meets the $x$-axis at $\mathrm{P}$ and $y$-axis at $\mathrm{Q}$, then $\mathrm{QA}: \mathrm{AP}$ is equal to :
If the circle $x^{2}+y^{2}-2 g x+6 y-19 c=0, g, c \in \mathbb{R}$ passes through the point $(6,1)$ and its centre lies on the line $x-2 c y=8$, then the length of intercept made by the circle on $x$-axis is :
Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2}-4 x-6=0$ and the ordinates of $\mathrm{P}$ and $\mathrm{Q}$ be the roots of $y^{2}+2 y-7=0$. If $\mathrm{PQ}$ is a diameter of the circle $x^{2}+y^{2}+2 a x+2 b y+c=0$, then the value of $(a+b-c)$ is _____________.
Consider three circles:
${C_1}:{x^2} + {y^2} = {r^2}$
${C_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}$
${C_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}$
If a line L : y = mx + c be a common tangent to C1, C2 and C3 such that C1 and C3 lie on one side of line L while C2 lies on other side, then the value of $20({r^2} + c)$ is equal to :
Let a triangle ABC be inscribed in the circle ${x^2} - \sqrt 2 (x + y) + {y^2} = 0$ such that $\angle BAC = {\pi \over 2}$. If the length of side AB is $\sqrt 2 $, then the area of the $\Delta$ABC is equal to :
Let the tangent to the circle C1 : x2 + y2 = 2 at the point M($-$1, 1) intersect the circle C2 : (x $-$ 3)2 + (y $-$ 2)2 = 5, at two distinct points A and B. If the tangents to C2 at the points A and B intersect at N, then the area of the triangle ANB is equal to :
If the tangents drawn at the points $O(0,0)$ and $P\left( {1 + \sqrt 5 ,2} \right)$ on the circle ${x^2} + {y^2} - 2x - 4y = 0$ intersect at the point Q, then the area of the triangle OPQ is equal to :
The set of values of k, for which the circle $C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$ lies inside the fourth quadrant and the point $\left( {1, - {1 \over 3}} \right)$ lies on or inside the circle C, is :
Let C be a circle passing through the points A(2, $-$1) and B(3, 4). The line segment AB s not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)^2} + {(y - 1)^2} = {{13} \over 2}$, then r2 is equal to :
A circle touches both the y-axis and the line x + y = 0. Then the locus of its center is :
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
$\therefore$ $h + k + r = 7$
$A = \{ (x,y) \in Z \times Z:{(x - 2)^2} + {y^2} \le 4\} $
$B = \{ (x,y) \in Z \times Z:{x^2} + {y^2} \le 4\} $
$C = \{ (x,y) \in Z \times Z:{(x - 2)^2} + {(y - 2)^2} \le 4\} $
If the total number of relation from A $\cap$ B to A $\cap$ C is 2p, then the value of p is :
C1 : x2 + y2 + 2y $-$ 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :
Then the minimum value of |r| such that $A \cup B \subseteq C$ is equal to
x2 + y2 $-$ 10x $-$ 10y + 41 = 0
x2 + y2 $-$ 22x $-$ 10y + 137 = 0
Circle M : x2 + y2 = 1
Circle N : x2 + y2 $-$ 2x = 0
Circle O : x2 + y2 $-$ 2x $-$ 2y + 1 = 0
Circle P : x2 + y2 $-$ 2y = 0
If the centre of circle M is joined with centre of the circle N, further center of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a :
