Question

Find the equation of the circle which passé through (-2,3),(4,5)and having centre on x axis

Answer 1

Step-by-step explanation:

Let center C be (0, y). Radius R satisfies: R^2 = 4 + (y-3)^2 = 16 + (y-5)^2.

==> 13 - 6y = 41 - 10y. ==> 4y = 28. → y = 7. C is (0, 7). R^2 = 20.

Equation of the desired circle is x^2 + (y - 7)^2 = 20.→ x^2 + y^2 - 14y + 29 = 0