The tangent to the circle x^2 + y^2 = 10 at the point (3,1) touches the circle x^2 + y^2 - 2 10 ,x - 20y + k = 0 , then the value of k is...

Step-by-step solution

The equation of the tangent to the circle x^2 + y^2 = 10 at the point (3,1) is given by T = 0 : 3x + y = 10 3x + y - 10 = 0 The equation of the second circle is x^2 + y^2 - 2 10 x - 20y + k = 0 . The center of this circle is ( 10 , 10) and its radius is r = ( 10 )^2 + (10)^2 - k = 110 - k . Since the line 3x + y - 10 = 0 touches the second circle, the perpendicular distance from the center to the line must be equal to the radius of the circle. The perpendicular distance d is: d = |3( 10 ) + 10 - 10| 3^2 + 1^2 = 3 1