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 $S$ , whose centre is $(2, - 3)$ , the radius of $S$ is

\sqrt{41}

unit

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3 \sqrt{5}

unit

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5 \sqrt{2}

unit

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2 \sqrt{5}

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Solution

The correct option is A $\sqrt{41}$ unit
$x^{2} + y^{2} + 4 x - 6 y - 12 = 0$
$C (2, - 3), r = \sqrt{4 + 9 + 12} = 5$
$C G = \sqrt{16 + 0} = 4$
${G P}^{2} = {C G}^{2} + {C P}^{2} = 16 + 25$
$C P = \sqrt{41}$

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Similar questions

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

S,

whose centre is

(2, - 3),

the radius of

S

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 of the diameter of the circle, given by the equation,

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)

5 \sqrt{3}

(B) 5

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 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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If one 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 (2,−3), the radius of S is

The correct option is A √41 unitx2+y2+4x−6y−12=0C(2,−3),r=√4+9+12=5CG=√16+0=4GP2=CG2+CP2=16+25CP=√41

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 $S$ , whose centre is $(2, - 3)$ , the radius of $S$ is

The correct option is A $\sqrt{41}$ unit
$x^{2} + y^{2} + 4 x - 6 y - 12 = 0$
$C (2, - 3), r = \sqrt{4 + 9 + 12} = 5$
$C G = \sqrt{16 + 0} = 4$
${G P}^{2} = {C G}^{2} + {C P}^{2} = 16 + 25$
$C P = \sqrt{41}$