The equation of the circle whose radius is 3 and which touches internally the circle x^2+y^2-4x-6y-12=0 at the point (-1 ,-1 ) is

Question

Accepted Answer

Correct answer: B. (b) Since, given equation of circle is, x^2+y^2-4x-6y-12=0 with centre C_1 (2 ,3 ) and radius r_i= 4+9+12=5. Let C_2 (h ,k ) and r=3 (given) for required circle touches the given circle at A (-1 ,-1 ). The point A (-1 ,-1 ) divides the line joining the centres C_1 (2 ,3 ) and C_2 (h ,k ) externally in 5 :3 so (-1 ,-1 )= ( 5h-62 , 5k-92 ) ⇒h= 45 and k= 75 so equation of required circle is (x- 45)^2+(y- 75)^2= (3 )^2 ⇒x^2+y^2- 8x5- 14y5+ 6525=9 ⇒5x^2+5y^2-8x-14y-32=0 Hence, option (b) is correct.

$The equation of the circle whose radius is 3 and which touches internally the circle x^{2} + y^{2} - 4 x - 6 y - 12 = 0 at the point (- 1, - 1) is$

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