Search any question & find its solution
Question:
Answered & Verified by Expert
If tangents are drawn to the circle $x^2+y^2=12$ at the points of intersection with the circle $x^2+y^2-5 x+3 y-2=0$, then the ordinate of the point of intersection of these tangents is
Options:
Solution:
1323 Upvotes
Verified Answer
The correct answer is:
$-\frac{18}{5}$
Let $(h, k)$ be the point of intersection of tangents then the chord of contact of tangents is the common chord of the circle $x^2+y^2=12$ and
$$
\begin{aligned}
x^2+y^2-5 x+3 y-2 & =0 \\
5 x-3 y-10 & =0
\end{aligned}
$$
i.e.
$$
5 x-3 y-10=0
$$
Also, the equation of the chord of with respect to the point is $h x+k y-12=0$
Equation $h x+k y-12=0$ and $5 x-3 y-10=0$ represents the same line.
$$
\because \quad \frac{h}{5}=\frac{k}{-3}=\frac{-12}{-10} \Rightarrow k=\frac{-18}{5}
$$
$$
\begin{aligned}
x^2+y^2-5 x+3 y-2 & =0 \\
5 x-3 y-10 & =0
\end{aligned}
$$
i.e.
$$
5 x-3 y-10=0
$$
Also, the equation of the chord of with respect to the point is $h x+k y-12=0$
Equation $h x+k y-12=0$ and $5 x-3 y-10=0$ represents the same line.
$$
\because \quad \frac{h}{5}=\frac{k}{-3}=\frac{-12}{-10} \Rightarrow k=\frac{-18}{5}
$$
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.