Search any question & find its solution
Question:
Answered & Verified by Expert
If the circle $S=x^2+y^2+2 g x+2 f y+c=0$ cuts each of the three circles $x^2+y^2+4 x+4 y+7=0, x^2+y^2-4 x+$ $4 y+7=0$ and $x^2+y^2-4 x-4 y+7=0$ orthogonally, then the equation of the tangent drawn at the point $(\sqrt{3}, 2)$ to the circle $\mathrm{S}=0$ is
Options:
Solution:
1923 Upvotes
Verified Answer
The correct answer is:
$\sqrt{3} \mathrm{x}+2 \mathrm{y}-7=0$.
$\begin{aligned} & \text {} 2 g(2)+2 f(2)=C+7 \\ & \Rightarrow \quad 4 g+4 f=C+7 \\ & 2 g(-2)+2 f(2)=C+7 \\ & \Rightarrow-4 g+4 f=C+7 \\ & 2 g(-2)+2 f(-2)=C+7 \\ & \Rightarrow \quad-4 g-4 f=C+7 \\ & (1)-(2), 8 g=0 \Rightarrow g=0 \\ & (2)+(3), C=-7 \\ & \therefore \quad f=0 \\ & \therefore \quad S: x^2+y^2-7=0 \\ & \quad T: x x_1+y y_1-7=0 \\ & \quad T \text { at }(\sqrt{3}, 2) \\ & \therefore \quad \sqrt{3} x+2 y=7=0 .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.