The equation of circle C is () A. (x−12)2+(y−3)2＝52B. (x−12)2+(y+3)2＝52C. (x+12)2+(y−3)2＝254D. (x+12)2+(y+3)2＝254

The equation of circle C is () A. (x−12)2+(y−3)2＝52B. (x−12)2+(y+3)2＝52C. (x+12)2+(y−3)2＝254D. (x+12)2+(y+3)2＝254

∵ the center of the circle is C (− 12,3), ∵ let the equation (x + 12) 2 + (Y − 3) 2 = R2 of the circle, and only one of the four options given is correct, that is, (x + 12) 2 + (Y − 3) 2 = 254, so choose C

It is known that the center coordinate of circle C is (- 1 / 2,3) and that circle C and line x + 2y-3 = 0 intersect at two points P and Q, and op ⊥ OQ and o are the origin of coordinates, so find circle C
Why does op ⊥ OQ have OP = OQ?
Why r = √ 10 / 2 and C: (x + 1 / 2) ^ 2 + (Y-3) ^ 2 = 5 / 2

OP and OQ are not equal. This problem requires the coordinates of m in PQ. Using the coordinates of circle C, PQ is a chord, so cm is perpendicular to PQ, so the slope of CM is 2, and the coordinates of M is (- 1,2), so PM = QM = om = √ 5, and cm = √ 5 / 2, so r = 5 / 2, so circle C is (x + 1 / 2) ^ 2 + (Y-3) ^ 2 = 25 / 4. According to the distance formula, the distance from any point on the circle to the center is equal, Because the distance has to be radical, the center equation is the square of R

It is known that the center coordinates of circle C are (- 1 / 2,3), and the circle C and the line x + 2y-3 = 0 intersect at two points P and Q, and op ⊥ OQ, O are the origin of coordinates, so the square of circle C is obtained
Why is the triangle OPQ isosceles triangle

The distance from the center of the circle to the straight line d = | - 1 / 2 + 6-3 | / √ 5 = √ 5 / 2;
From op ⊥ OQ (with OP = OQ), r = √ 2D = √ 10 / 2 can be deduced
Circle C: (x + 1 / 2) ^ 2 + (Y-3) ^ 2 = 5 / 2