Circle Ques 48
- If the circle $x^{2}+y^{2}+2 x+2 k y+6=0$ and $x^{2}+y^{2}+2 k y+k=0$ intersect orthogonally, then $k$ is
(a) 2 or $-3 / 2$
(b) -2 or $-3 / 2$
(c) 2 or $3 / 2$
(d) -2 or $3 / 2$
$(2000,2 M)$
Show Answer
Answer:
Correct Answer: 48.(a)
Solution:
Formula:
Since the given circles intersect orthogonally.
$ \begin{alignedat} & \therefore & 2(1)(0)+2(k) & (k)=6+k \\ & & & {\left[\because 2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2}\right] } \\ & & & \\ & & 2 k^{2}-k-6=0 \Rightarrow & k=-\frac{3}{2}, 2\frac{2}{3} \end{aligned} $
BETA
