Circle Ques 49
- The common tangent to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}+6 x+8 y-24=0$ also passes through the point
(2019 Main, 9 April II)
(a) $(6,-2)$
(b) $(4,-2)$
(c) $(-6,4)$
(d) $(-4,6)$
Show Answer
Answer:
Correct Answer: 49.(a)
Solution:
Formula:
Common Tangents of Two Circles
- Given circles are $x^{2}+y^{2}=4$, centre $c _1(0,0)$ and radius $r _1=2$
and $x^{2}+y^{2}+6 x+8 y-24=0$, centre $c _2(-3,-4)$ and radius $r _2=7$
$\because \quad c _1 c _2=\sqrt{9+16}=5$ and $\left|r _1-r _2\right|=5$
$\because \quad c _1 c _2=\left|r _1-r _2\right|=5$
$\therefore$ circle $x^{2}+y^{2}=4$ touches the circle $x^{2}+y^{2}+6 x+8 y-24=0$ internally.
So, equation of common tangent is
$$ S _1-S _2=0 $$
$$ \Rightarrow \quad 6 x+8 y-20=0 $$
$\Rightarrow \quad 3 x+4 y=10$
The common tangent passes through the point $(6,-2)$, from the given options.
BETA
