Search any question & find its solution
Question:
Answered & Verified by Expert
If the circles $x^{2}+y^{2}+2 x+2 k y+6=0$, $x^{2}+y^{2}+2 k y+k=0$ intersect orthogonally, then $k$ is
Options:
Solution:
1961 Upvotes
Verified Answer
The correct answer is:
2 or $-3 / 2$
Since, circles cut orthogonally, if
$$
\begin{aligned}
& & 2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime} \\
\Rightarrow & & \quad 0+2 k^{2}=6+k \\
\Rightarrow & & 2 k^{2}-k-6=0 \\
\Rightarrow & &(2 k+3)(k-2)=0
\end{aligned}
$$
$\Rightarrow \quad k=2$ or $-3 / 2$
$$
\begin{aligned}
& & 2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime} \\
\Rightarrow & & \quad 0+2 k^{2}=6+k \\
\Rightarrow & & 2 k^{2}-k-6=0 \\
\Rightarrow & &(2 k+3)(k-2)=0
\end{aligned}
$$
$\Rightarrow \quad k=2$ or $-3 / 2$
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.