A point P moves such that the sum of the squares of its distances from the points A(-14, -7) and B(-6, -3) is 90 . The locus of P is a circle C . If tangents are drawn from the origin O to the circle C touching it at points Q and R , then the area of the triangle OQR is

Step-by-step solution

Let P(x, y) be the moving point. Given, PA^2 + PB^2 = 90 (x+14)^2 + (y+7)^2 + (x+6)^2 + (y+3)^2 = 90 x^2 + 28x + 196 + y^2 + 14y + 49 + x^2 + 12x + 36 + y^2 + 6y + 9 = 90 2x^2 + 2y^2 + 40x + 20y + 290 = 90 2x^2 + 2y^2 + 40x + 20y + 200 = 0 x^2 + y^2 + 20x + 10y + 100 = 0 This represents the circle C . Its center is (-10, -5) and its radius R is: R = 10^2 + 5^2 - 100 = 25 = 5 The length of the tangent L from the origin (0,0) to the circle C is: L = S₁ = 0^2 + 0^2 + 20(0) + 10(0) + 100 = 100 = 10 The area of the tria