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 :

5 \sqrt{2}

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5 \sqrt{3}

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5

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Solution

The correct option is A $5 \sqrt{3}$
Given circle is $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$

Center is $(2, - 3)$ , radius is $5$ units

Let $r$ be the required radius. $(- 3, 2)$ is the center of that circle. So, distance
between two centres will be $\sqrt{(- 3 - 2)^{2} + (2 + 3)^{2}} = 5 \sqrt{2}$

Now,

$r^{2} = 5^{2} + (5 \sqrt{2})^{2} = 25 + 50$

Radius is $r = \sqrt{75} = 5 \sqrt{3}$

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Similar questions

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:

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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

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x^{2} + y^{2}

-4x +6y - 12 = 0, is a chord of a circle S, whose centre is at ( -3,2), then the radius of S is:
(A)

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Q. 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

Q. If one 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

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whose centre is

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the radius of

S

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If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0 is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is :

The correct option is A 5√3Given circle is x2+y2−4x+6y−12=0Center is (2,−3), radius is 5 unitsLet r be the required radius. (−3,2) is the center of that circle. So, distance between two centres will be √(−3−2)2+(2+3)2=5√2Now, r2=52+(5√2)2=25+50Radius is r=√75=5√3

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 :