The lengths of tangents from a fixed point to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the coaxial system of circles be given by $x^2 + y^2 + 2g_i x + c = 0$ for $i = 1, 2, 3$,where $g_i$ are variables and $c$ is a constant.Let t

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let the coaxial system of circles be given by $x^2 + y^2 + 2g_i x + c = 0$ for $i = 1, 2, 3$,where $g_i$ are variables and $c$ is a constant.Let the fixed point be $(h, k)$. The square of the length of the tangent from $(h, k)$ to the $i$-th circle is $t_i^2 = h^2 + k^2 + 2g_i h + c$.The centers of the circles are $P(-g_1, 0)$,$Q(-g_2, 0)$,and $R(-g_3, 0)$.The distances between the centers are $QR = |g_3 - g_2|$,$RP = |g_1 - g_3|$,and $PQ = |g_2 - g_1|$.Consider the expression $QRt_1^2 + RPt_2^2 + PQt_3^2 = (g_3 - g_2)(h^2 + k^2 + 2g_1 h + c) + (g_1 - g_3)(h^2 + k^2 + 2g_2 h + c) + (g_2 - g_1)(h^2 + k^2 + 2g_3 h + c)$.Expanding this,we get $(h^2 + k^2 + c)(g_3 - g_2 + g_1 - g_3 + g_2 - g_1) + 2h(g_1 g_3 - g_1 g_2 + g_2 g_1 - g_2 g_3 + g_3 g_2 - g_3 g_1) = 0 + 0 = 0$.

The lengths of tangents from a fixed point to three circles of a coaxial system are $t_1, t_2, t_3$. If $P, Q,$ and $R$ are the centers of these circles,then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

Similar Questions

If the circle $S=0$ cuts the circles $x^2+y^2-2x+6y=0$,$x^2+y^2-4x-2y+6=0$,and $x^2+y^2-12x+2y+3=0$ orthogonally,then the equation of the tangent at $(0,3)$ on $S=0$ is

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{3}$,then a value of $k$ is

If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to

If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ intersect orthogonally,then the value of $k$ is:

If a circle has its centre on the line $x-y-1=0$ and passes through the points of intersection of the two circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y-7=0$,then the centre of that circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module