If the circles x^2+y^2-2x-2y+k=0 and x^2+y^2+4x+6y+4=0 touch each other externally, then the point of contact of the two circles is

Question

Accepted Answer

Correct answer: A. (a) Given two circles are x^2+y^2-2x-2y+k=0 and x^2+y^2+4x+6y+4=0 Touching each other externally, so c_1c_2=r_1+r_2 where, c_1 (1 ,1 ) ,c_2 (-2 ,-3 ) and r_1= 2-k ,r_2=3 So, 9+16= 2-k+3 ⇒2= 2-k ⇒4=2-k ⇒k=-2 The point of contact P of two circles, divides the line joining centres c_1 and c_2 in the ratio r_1 :r_2 ∴P ( 3-45 , 3-65 )=P(- 15 , -35)

$If the circles x^{2} + y^{2} - 2 x - 2 y + k = 0 and x^{2} + y^{2} + 4 x + 6 y + 4 = 0 touch each other externally, then the point of contact of the two circles is$

Solution

More Parabola PYQs