equation of a circle in the middle of two outer touching circles

Question

I am asked:

Three circles touch externally as shown in the diagram (The diagram shows a large circle sandwiched in between 2 smaller circles. The centres are colinear and the equations of the two smaller circles are:

${x^2 + y^2 + 20 x -16y - 139 = 0}$

${x^2 + y^2 - 28x + 48y + 747 = 0}$

Find the equation of the large circle.

My logic is that because they are externally touching, I can find the distance between the mid point of the two outer circles. I can then subtract the 2 radii from distance to find the radius of the big circle. I can find the midpoint of the larger circle by finding the midpoint between the two outer smaller circle's midpoints.

I factorised the 2 smaller circle equations to be:

${(x + 10) + (^2 + (y-8)^2} = 25$

And

${(x - 14)^2 + (y + 24)^2 = 25}$

So the 2 circles have the coordinates and radius of:

(-10, 8) r = 5 and (14, -24) r = 5

The distance between the 2 midpoints is:

${\sqrt{{(14 -- 10) ^2 + (-24 - 8)^2}}}$ = 40

That would make the diameter of the larger circle 40 - the 2 radii which is 30 so I am saying the radius of the larger circle is 15.

I found the midpoint between the 2 points to be:

${({-10 + 14 \over 2}), ({8 + -24\over 2})} = (2, 8)$

So I would now construct the radius of the circle from the coordinates (2, 8) and a radius of 15.

Which would give:

${x^2 - 4x + y^2 -16y -225 = 0}$

But the answer in the book is

${x^2 - 4x + y^2 -16y -157 = 0}$

I have absolutely no idea where -157 would come from.

Answer 1

You were nearly there.

So I would now construct the radius of the circle from the coordinates (2, 8) and a radius of 15.

$(x - 2)^2 + (y - 8)^2 = 15^2$

$x^2 - 4x + 4 + y^2 - 16y + 64 = 225$

$x^2 - 4x + y^2 - 16y = 225 - 64 - 4 = 157$

You just forgot the squares of the constants when you expanded your x and y terms.