Find the equation of a circle with two given points through which circle passes.

Question

Today, I was solving a problem which is described below.

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

I tried to solve and found the intersection points of both ellipses which are $(2,-1)$ and $(2,1)$. But the problem is that how could I show the required result for the question?

I haven't solved any equation of circle with just two given point lying on the circle. Please help me solving it.

Answer 1

Note that center will lie on $x-$axis, therefore let the required equation of circle be of type

$(x-a)^2+y^2=r^2\implies x^2-2ax+a^2+y^2=r^2$

Now it passes through $(2,1) $ and $(2,-1)$, therefore these points will satisfy the equation above

$4-4a+a^2+1=r^2\implies5-4a+a^2=r^2$

Substitute $r^2$ back to get required result

Also note that $a$ here is $x-$ coordinate of center of circle.

Since you want to know that How to show that center will be at $x-$ axis, I will tell you two ways-

--> Take any chord of a circle. If the reflection of any point on it lies on the circle, then only that chord is diameter. Here I used converse of same thing. Since $(2,1)$ and $(2,-1)$ lie on a circle and are reflection of each other on $x-$ axis, therefore $x-$ axis should coincide with one of diameters of circle

--> Let center be $(\alpha,\beta)$ . Since two points lie on circle, therefore their distance from the center should be same which gives $(\alpha-2)^2+(\beta-1)^2=(\alpha-2)^2+(\beta+1)^2$ , solving which gives $\beta=0$ and thus center lies on $x-$ axis.

Answer 2

HINT:

You have an equation of one circle $x^2 + y^2 = 5$ that passes through both points, and now add to that a multiple of the equation of the line through both points.

Answer 3

Let $ A(2,-1) $ and $ B(2,1) $.

if $ C(a,b) $ is the center, then

$$AC^2=BC^2 \implies$$

$$(a-2)^2+(b+1)^2=$$ $$(a-2)^2+(b-1)^2$$ $$\implies b=0$$ $$\implies C \text{ is in } x- \text{axis}.$$

the equation of the circle will be of the form

$$(x-a)^2+y^2=AC^2$$ $$=(a-2)^2+(0+1)^2$$

or $$x^2-2ax+y^2=5-4a$$

Answer 4

Method without solving the equations:

$(x + 2)^2 + 2y^2 = 18$ , $9(x - 1)^2 + 16y^2= 25$

There is a fast way to solve this question... but it requires a little cleverness:

$$ \lambda \left[ (x+2)^2 + 2y^2 -18\right] + \mu \left[ 9(x-1)^2 + 16y^2 -25\right] =0 \tag{1}$$

We want:

$$ \lambda +9 \mu =2 \lambda + 16 \nu \tag{1.5}$$

Which means, $-7 \mu = \lambda$, substituting in equation (1), we have:

$$ -7\mu \left[ (x+2)^2 + 2y^2 -18\right] + \mu \left[ 9(x-1)^2 + 16y^2 -25\right] =0 \tag{2}$$

By algebra, we find (2) becomes:

$$x^2 -23 x + y^2 + 41=0$$