A pair of tangents are drawn to the circle x^2+y^2+6 x-4 y-12=0 from a point P (-4,-5) , then the area enclosed between these tangents and the area of the circle is

Step-by-step solution

Given the circle equation x^2 + y^2 + 6x - 4y - 12 = 0 , comparing with the general form x^2 + y^2 + 2gx + 2fy + c = 0 yields g = 3 , f = -2 , and c = -12 . The center is C(-g, -f) = (-3, 2) and the radius is r = g^2 + f^2 - c = 9 + 4 + 12 = 5 . For external point P(-4, -5) , the distance to center is PC = (-4 + 3)^2 + (-5 - 2)^2 = 1 + 49 = 5 2 . The tangent length from P is PC^2 - r^2 = 50 - 25 = 5 . In right triangle PCT , with legs CT = 5 and PT = 5 , CPT = 45^ and PCT = 45^ . The angle between tangents is 2 45^