The topic of compulsory mathematics in senior one and compulsory mathematics in senior two.

1. Find the center of a straight line X-Y-4=0, and cross the circle X? +Y？ +6X-4=0 and circle x? +Y？ Equation of the circle at the intersection of +6Y-28=0

Solution: circle x? +Y？ +6X-4=0 can be converted into (X+3)? +Y？ = 13, center of the circle (-3,0)

Circle x? +Y？ +6Y-28=0 can be converted into x? +(Y+3)？ =37, center of the circle (0, -3)

Because the circle you want passes through the circle x? +Y？ +6X-4=0 and circle x? +Y？ +6Y-28=0, so the center of the circle is on the straight line passing through points (-3,0) and (0,3), that is, the straight line X+Y=-3.

And the center of the circle is on the straight line X-Y-4=0.

So the coordinates of the center of the circle are (1/2, -7/2).

X round again? +Y？ +6X-4=0 and circle x? +Y？ The intersection of +6Y-28=0 is (-1, 3) and (-6, 2).

So the square of the radius of the circle =(- 1- 1/2)? +(3+7/2)? =89/2

So the equation for finding a circle is (X- 1/2)? +(Y+7/2)？ =89/2

2. Find the point M(2, -2) and the circle x? +Y？ -6X=0 and x? +Y？ Equation of Circle with Intersection =4

Solution: circle x? +Y？ -6X=0 and x? +Y？ =4 The coordinates of intersection point are (2/3, 4√2/3), (2/3, -4√2/3).

Let the equation of a circle be (X-a). +(Y-b)？ =r？

So: (2-a)? +(-2-b)？ =r？

(2/3-a)？ +(4√2/3-b)？ =r？

(2/3-a)？ +(-4√2/3-b)？ =r？

Therefore: b=0, a=3/2, r? = 17/4