Circle 2 Question 5
6. If one of the diameters of the circle, given by the equation, $x^{2}+y^{2}-4 x+6 y-12=0$, is a chord of a circle $S$, whose centre is at $(-3,2)$, then the radius of $S$ is
(a) $5 \sqrt{2}$
(b) $5 \sqrt{3}$
(c) 5
(d) 10
(2016 Main)
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Answer:
Correct Answer: 6. (c)
Solution:
- Given equation of a circle is $x^{2}+y^{2}-4 x+6 y-12=0$, whose centre is $(2,-3)$ and radius
$$ =\sqrt{2^{2}+(-3)^{2}+12}=\sqrt{4+9+12}=5 $$
Now, according to given information, we have the following figure.
$$ x^{2}+y^{2}-4 x+6 y-12=0 $$
Clearly, $A O \perp B C$, as $O$ is mid-point of the chord.
Now, in $\triangle A O B$, we have
and
$$ \begin{aligned} O A & =\sqrt{(-3-2)^{2}+(2+3)^{2}} \\ & =\sqrt{25+25}=\sqrt{50}=5 \sqrt{2} \end{aligned} $$
$$ \therefore \quad A B=\sqrt{O A^{2}+O B^{2}}=\sqrt{50+25}=\sqrt{75}=5 \sqrt{3} $$
