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. (b)

Solution:

  1. 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} $



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