if the equation of the circle is (-3,3) and (9,9) as 2 endpoints of a diameter is expanded as x^2+y^2+Ax+By=0?

Question

then a= and b=

can someone show me how to do this

Answer 1

For a circle of radius R with a center at the origin, the equation is:

x^2 + y^2 = R^2

If the origin is at (a,b) the equation becomes:

(x-a)^2 + (y-b)^2 = R^2

Multiplying this out you get:

x^2 - 2ax + a^2 + y^2 - 2by + b^2 = R^2 or

x^2 - 2ax + y^2 - 2by + (a^2 + b^2 - R^2) = 0

so A = -2a, B = -2b

So the problem becomes finding a and b.

But if the line PQ is a diameter of the circle, then the point half way between P and Q is the center.

You have the coordinates of P and Q so you can compute the coordinates of the center, a and b.

That gives you the values of A and B that you need.