A circle is a locus of all the points that are at a same distance which is measured from a fixed point. Orthogonal is the condition of perpendicularity.

So, the circles that cut each other at right angles are known as orthogonal circles. This implies orthogonal circles are perpendicular to each other. When we talk about angles between the circles we actually mean the angles between the tangents to the circle at the point of intersections.

Orthogonal Circles Proof

When we have two circles in a plane which are intersecting, then we will first show that orthogonal circle intersect at exactly two points

CASE 1:
Consider that the two circles have no point or zero point in common. Then we say that the given circles are disjoint as well as trivial.

CASE 2:
The given circles have exactly and only one point in common. Then this implies that the circles are tangent to each other as they only touch each other. In such case both the circles that are tangent to each other have a common tangent line at the point they are tangent to each other and hence they cannot be perpendicular to each other.

CASE 3:
Consider that the two circles are having three or more points in common. In this case they are the same circle only. Also these three points cannot be collinear as well and hence they will surely form a triangle and that both the circles will circumscribe the triangle.

CASE 4:
Now, we are left with the condition that the circles are intersecting at exactly two points. Now if the circles are intersecting they may be orthogonal but in all other cases they cannot be as explained above.

Orthogonal Circles Condition

If the two circles have radii r, R and their centers are located at ‘d’ distance apart from each other, then we say that the circle are orthogonal if and only if
R$^{2}$ + r$^{2}$ = d$^{2}$

This can be rewritten as if the coordinates of the centers are (a, b) and (c, d),

R$^{2}$ + r$^{2}$ = (a - c)$^{2}$ + (b - d)$^{2}$ …..(i)

Basically we used the distance formula between the centers that is equal to ‘d’ for cases where ‘d’ is not known directly.

In the Cartesian system circles with equations,
x$^{2}$ + y$^{2}$ + 2 f x + 2 g y + c = 0

x$^{2}$ + y$^{2}$ + 2 f’x + 2 g’ y + c’ = 0

Are said to be orthogonal if and only if
2 g g’ + 2 f f’ = c + c’

Equation of Orthogonal Circles

From equation (i) above we have R$^{2}$ + r$^{2}$ = (a - c)$^{2}$ + (b - d)$^{2}$ where R is the radius of one circle with center (a, b) and ‘r’ is the radius of another circle with center (c, d).

From this equation we can deduce that the center of one of the circles is always lying outside its orthogonal circle. Assume that (m, n) is any

arbitrary point outside the circle (x - h)$^{2}$ + (y - k)$^{2}$ = r$^{2}$, then we can at any time draw a circle with center (m, n) which will be

orthogonal to the circle (x - h)$^{2}$ + (y - k)$^{2}$ = r$^{2}$. In such case, the square of the limited tangent from the point (m, n) is always equal

to the power of the point taken with respect to the circle (x - h)$^{2}$ + (y - k)$^{2}$ = r$^{2}$ and hence is (m - h)$^{2}$ + (n - k)$^{2}$ r$^{2}$.

Then the equation of the circle orthogonal to the circle (x - h)$^{2}$ + (y - k)$^{2}$ = r$^{2}$ and with center (m, n) is given by:

(x - m)$^{2}$ + (y - n)$^{2}$ = (m - h)$^{2}$ + (n - k)$^{2}$ - r$^{2}$

More generalized form of this equation can be when we take arbitrary point as (x_0, y_0) and center of the given circle to be (a, b). Then we rewrite it as
(x - x_0)$^{2}$ + (y - y_0)$^{2}$ = (x_0 - a)$^{2}$ + (y_0 - b)$^{2}$ - r$^{2}$

Let us see an example on orthogonal circle to understand it better.

Example:

Given two circle x$^{2}$ + y$^{2}$ + 2x - 2y + 1 = 0 and x$^{2}$ + y$^{2}$  + 4x + 4y + 3 = 0. Find the equation of the circle that is orthogonal to both these circle with center on x - y = 1

Solution:

x$^{2}$ + y$^{2}$ + 2x - 2y + 1 = 0                                 …..(i)

x$^{2}$ + y$^{2}$ + 4x + 4y + 3 = 0                                …..(ii)

Let the required equation be x$^{2}$ + y$^{2}$ + 2fx + 2gy + c = 0     …..(iii)

Since it is orthogonal to both (i) and (ii) we have

2g - 2f = c + 1

And 4g + 4f = c + 3

Also (-g, -f) lies on x - y = 1

$\Rightarrow$ g - f = -1
$\Rightarrow$ c = -3
$\Rightarrow$ g = $\frac{-1}{2}$
$\Rightarrow$ f = $\frac{1}{2}$

Therefore, required equation is x$^{2}$ + y$^{2}$ + x - y - 3 = 0