Search any question & find its solution
Question:
Answered & Verified by Expert
3. The equation of the circle which intersects circles $x^2+y^2+x+2 y+3=0, x^2+y^2+2 x+4 y+5=0$ and $x^2+y^2-7 x-8 y-9=0$ at right angle, will be
Options:
Solution:
2977 Upvotes
Verified Answer
The correct answer is:
$3\left(x^2+y^2\right)+4(x+y)-3=0$
Let circle be $x^2+y^2+2 g x+2 f y+c=0$. Then according to the conditions given,
$g+2 f=c+3...(i)$
$\begin{aligned}& 2 g+4 f=c+5 ...(ii)\\& -7 g-8 f=c-9 ...(iii)\\& \Rightarrow g=\frac{2}{3}, f=\frac{2}{3}, c=-1\end{aligned}$
Therefore, the required equation is
$3\left(x^2+y^2\right)+4(x+y)-3=0$
$g+2 f=c+3...(i)$
$\begin{aligned}& 2 g+4 f=c+5 ...(ii)\\& -7 g-8 f=c-9 ...(iii)\\& \Rightarrow g=\frac{2}{3}, f=\frac{2}{3}, c=-1\end{aligned}$
Therefore, the required equation is
$3\left(x^2+y^2\right)+4(x+y)-3=0$
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.