write the equation of circle containing point p(0,-3) q(4,0) whose center x+2y=0

Question

The equation of circle

Answer 1

General equation of circle is (x-h)^2+(y-k)^2=a^2 where a is radius and (h,k) is centre.

At point p: h^2+(k+3)^2=a^2 and at point q: (4-h)^2+k^2=a^2.

h^2+k^2+6k+9=h^2-8h+16+k^2=a^2; 6k+9=16-8h, k=(7-8h)/6. Centre is (h,(7-8h)/6).

At centre, x+2y=0; h+(7-8h)/3=0; 3h+7-8h=0; 5h=7; h=1.4. Therefore k=(7-11.2)/6=-4.2/6=-0.7, and centre is (1.4,-0.7). a^2=1.96+5.29 or 6.76+0.49=7.25.

Equation of circle is: (x-1.4)^2+(y+0.7)^2=7.25;

x^2-2.8x+1.96+y^2+1.4y+0.49=7.25;

x^2-2.8x+y^2+1.4y=4.8; 5x^2-14x+5y^2+7y=24 are alternative ways of writing the equation.

CHECK: When x=0 y should be -3: 5*9-21=24, OK; when y=0 x should be 4: 5*16-56=24, OK.