Question

Equation of the circle passing through (-7, 1) and having centre at (-4,-3) is​

Answer 1

Centre of the circle is (3,4) = (h,k)

The circle passes through (-2,1). So the radius of the circle r is nothing but distance between (3,4) and (-2,1).

So radius r = √{(x2-x1)^2 + (y2-y1)^2}

= √{(-2–3)^2+(1–4)^2} = √(25+9 =√34.

Now equation of the circle with centre (h,k) and radius r is (x-h)^2 + (y-k)^2 = r^2

=> (x-3)^2 + (y-4)^2 = (√34)^2

=> x^2 - 6x + 9 + y^2 - 8y + 16 = 34. On simplification we get

x^2+y^2 -6x-8y-9=0

hope it is helpfull to you