It is known that the center coordinate of circle C is (- 1,3), and the circle and the straight line x + Y-3 = 0 intersect at P and Q, and OP is perpendicular to OQ, O is the origin of coordinates, so the equation of circle C is obtained

It is known that the center coordinate of circle C is (- 1,3), and the circle and the straight line x + Y-3 = 0 intersect at P and Q, and OP is perpendicular to OQ, O is the origin of coordinates, so the equation of circle C is obtained

Let (x + 1) ^ 2 + (Y-3) ^ 2 = R ^ 2, x + Y-3 = 02x ^ 2 + 2x + 1-r ^ 2 = 0. Weida theorem: XP + XQ = - 1 XP * XQ = (1-r ^ 2) / 2 and op perpendicular to OQ, then (YP / XP) * (YQ / XQ) = - 1 ((3-xp) / XP) * ((3-xq) / XQ) = - 19-3xp-3xq + xpxq -------- = - 1XP * XQ is substituted into XP + XQ = - 1 XP

It is known that the center of circle C is on the straight line L: x-2y-1 = 0 and passes through the origin and a (2,1). The standard equation of circle C is obtained

Let the standard equation of the circle be (x-a) 2 + (y-b) 2 = R2, the center of the circle (a, b), and the radius R. ∵ the center of the circle C is on the straight line L: x-2y-1 = 0, and through the origin and a (2, 1), ∵ a − 2B − 1 = 0a2 + B2 = R2 (2 − a) 2 + (1 − b) 2 = R2, a = 65B = 110r2 = 2920 is obtained

The equation of the circle whose center is (0, - 2) and tangent to x-axis and origin

Draw a picture. The center of the circle is (0, - 2) and the radius is 2, so the equation x ^ 2 + (Y-2) ^ 2 = 2