How to find the equation about circle The equation of circle tangent to X-axis with point C (2-3) as its center is? （X-a）²+（Y-b）²=R²

How to find the equation about circle The equation of circle tangent to X-axis with point C (2-3) as its center is? （X-a）²+（Y-b）²=R²

C（2,-3）
So (X-2) 2 + (y + 3) 2 = R2
It is tangent to the X axis, so r = | - 3 | = 3
The equation is: (X-2) 2 + (y + 3) 2 = 9

La-b-1 | / radical 2 = R, c2-a) square + c-1-b) square = R, square 2A + B = O, this is the problem solving, to be very detailed

If the center of the circle is on the line 2x + y = 0, find the equation of the circle. From 2A + B = 0, get b = - 2A, substitute la-b-1 | / √ 2 = R and square, get (3a-1) ^ 2 = 2R ^ 2, substitute (2-A) ^ 2 + (- 1 -)

The equation of the straight line of the two adjacent sides of the parallelogram is x + y + 1 = 0 and 3x-4 = 0. The intersection point of its diagonal is d (3,3). The equation of the straight line of the other two sides is obtained

From the meaning of the problem, we get x + y + 1 = 03x − y + 4 = 0, and the solution is x = − 54y = 14, that is, the vertex of two adjacent sides of a parallelogram is (− 54,14), and the diagonal intersection is d (3,3), then the other vertex on the diagonal is (294234). ∵ the other two sides of the straight line are parallel to the straight line x + y + 1 = 0 and 3x-y + 4 = 0, and their slopes are - 1 and 3, that is, their equation is y-234 = - (x − 294), And y-234 = 3 (x − 294), in addition, the linear equations of the two sides are x + y-13 = 0 and 3x-y-16 = 0, respectively