The radius of the circle centred at $(4, 5)$ and passing through the centre of the circle $x^{2} + y^{2} + 4 x + 6 y - 12 = 0$ is

2 \sqrt{5}

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2 \sqrt{10}

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3 \sqrt{5}

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3 \sqrt{10}

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Solution

The correct option is A $2 \sqrt{5}$
Let the center $A (4, 5)$
$x^{2} + y^{2} + 4 x + 6 y - 12 = 0$
On comparing with it
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$g = 2, f = 3, c = - 12$
center of the circle $= (- g, - f) = (- 2, - 3)$
Then required radius of the circle \
$= \sqrt{(4 + 2)^{2} + (5 + 3)^{2}} = 10$

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The radius of the circle centred at (4,5) and passing through the centre of the circle x2+y2+4x+6y−12=0 is

The correct option is A 2√5Let the center A(4,5)x2+y2+4x+6y−12=0On comparing with itx2+y2+2gx+2fy+c=0g=2,f=3,c=−12center of the circle =(−g,−f)=(−2,−3)Then required radius of the circle \=√(4+2)2+(5+3)2=10

The radius of the circle centred at $(4, 5)$ and passing through the centre of the circle $x^{2} + y^{2} + 4 x + 6 y - 12 = 0$ is