Show that the equation of any circle passing through the poi

Show that the equation of any circle passing through the points of intersection of the ellipse
(x + 2)² + 2y² = 18
and the ellipse
9(x?1)² + 16y² = 25
can be written in the form
x² ?2ax + y² = 5?4a .

Solution

First we have to ?nd the intersections of the two ellipses by solving the simultaneous equations
(x + 2)² + 2y² = 18

9(x-1)² + 16y² = 25 .

We can eliminate y by multiplying the ?rst equation by 8 and subtracting the two equations:

8(x + 2)² -9(x-1)² = 144-25 i.e. x² ?50x + 96 = 0 i.e. (x-2)(x-48) = 0 .

The two possible values for x at the intersections are therefore 2 and 48.

Next we ?nd the values of y at the intersection. Taking x = 2 and substituting into the ?rst simultaneous equation, we have 16 + 2y² = 18, so y = +-1.

Taking x = 48 gives 50² + 2y² = 18 which has no (real) roots. Thus there are two points of intersection, at (2,1) and (2,-1).

Suppose a circle through these points has centre (p,q) and radius R.

Then the equation of the circle is (x-p)² + (y-q)² = R². Setting (x,y) = (2,1) and (x,y) = (2,-1) gives two equations:.

(2-p)² + (1-q)² = R² , (2-p)2 + (-1-q)² = R² .

Subtracting gives q = 0 (it is obvious anyway that the centre of the circle must lie on the y axis).

Thus the equation of any circle passing through the intersections is (x-p)² + y2 = (2-p)² + 1 , which simpli?es to the given result with p = a.