Search any question & find its solution
Question:
Answered & Verified by Expert
The locus of the centre of circles passing through $(a, b)$ and cut the circle $x^2+y^2-2 x+4 y-4=0$ orthogonally is
Options:
Solution:
2772 Upvotes
Verified Answer
The correct answer is:
$(a-1) x+(b+2) y=\left(\frac{a^2+b^2+4}{2}\right)$
Let the equation of one of the circles be
$$
x^2+y^2+2 g x+2 f y+c=0
$$
Since, it passes through $(a, b)$.
Hence, $a^2+b^2+2 g a+2 f b+c=0$
Also, it cuts the circle $x^2+y^2-2 x+4 y-4=0$ orthogonally.
$$
\begin{aligned}
& \text { Then } \quad-2 g(\mathrm{l})-2 f(-2)=c+(-4) \\
& \Rightarrow \quad-2 g+4 f=c-4 \\
& \Rightarrow \quad-2 g+4 f=-a^2-b^2-2 g a-2 f b-4 \\
& \Rightarrow \quad a^2+b^2+(-2+2 a) g \\
& +(4+2 b) f+4=0 \\
&
\end{aligned}
$$
Thus, the locus of the centre $(-g,-f)$ is
$$
\begin{aligned}
& \left(a^2+b^2+4\right)+2(-1+a) x+2(2+b) y=0 \\
& \Rightarrow \quad(a-1) x+(b+2) y=-\left(\frac{a^2+b^2+4}{2}\right)
\end{aligned}
$$
$$
x^2+y^2+2 g x+2 f y+c=0
$$
Since, it passes through $(a, b)$.
Hence, $a^2+b^2+2 g a+2 f b+c=0$
Also, it cuts the circle $x^2+y^2-2 x+4 y-4=0$ orthogonally.
$$
\begin{aligned}
& \text { Then } \quad-2 g(\mathrm{l})-2 f(-2)=c+(-4) \\
& \Rightarrow \quad-2 g+4 f=c-4 \\
& \Rightarrow \quad-2 g+4 f=-a^2-b^2-2 g a-2 f b-4 \\
& \Rightarrow \quad a^2+b^2+(-2+2 a) g \\
& +(4+2 b) f+4=0 \\
&
\end{aligned}
$$
Thus, the locus of the centre $(-g,-f)$ is
$$
\begin{aligned}
& \left(a^2+b^2+4\right)+2(-1+a) x+2(2+b) y=0 \\
& \Rightarrow \quad(a-1) x+(b+2) y=-\left(\frac{a^2+b^2+4}{2}\right)
\end{aligned}
$$
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.