# Angle between two Circles

## Angle between two Circles

Statement: If θ is the angle between the circles x² + y² + 2gx + 2fy + c = 0, x² + y² + 2g’x + 2f’y + c = 0 then $$\cos \theta =\frac{c+c’-2(gg’+ff’)}{2\sqrt{{{g}^{2}}+{{f}^{2}}-c}\sqrt{g{{‘}^{2}}+f{{‘}^{2}}-c’}}$$.

Proof: Let C₁, C₂ be the centre S and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively and C₁C₂ = d.

∴ C₁ = (-g, -f), C₂ = (-g’, -f’),

$${{r}_{1}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c},\,\,{{r}_{2}}=\sqrt{g{{‘}^{2}}+f{{‘}^{2}}-c’}$$,

Now $$\cos \theta =\frac{{{d}^{2}}-r_{1}^{2}-r_{2}^{2}}{2{{r}_{1}}{{r}_{2}}}=\frac{{{\left( g-g’ \right)}^{2}}+{{\left( f-f’ \right)}^{2}}-\left( {{g}^{2}}+{{f}^{2}}-c \right)-\left( g{{‘}^{2}}+f{{‘}^{2}}-c’ \right)}{2\sqrt{{{g}^{2}}+{{f}^{2}}-c}\sqrt{g{{‘}^{2}}+f{{‘}^{2}}-c’}}$$,

$$=\frac{{{g}^{2}}+g{{‘}^{2}}-2gg’+{{f}^{2}}+f{{‘}^{2}}-2ff’-{{g}^{2}}-{{f}^{2}}+c-g{{‘}^{2}}-f{{‘}^{2}}+c’}{2\sqrt{{{g}^{2}}+{{f}^{2}}-c}\sqrt{g{{‘}^{2}}+f{{‘}^{2}}-c’}}$$,

$$=\frac{c+c’-2\left( gg’+ff’ \right)}{2\sqrt{{{g}^{2}}+{{f}^{2}}-c}\sqrt{g{{‘}^{2}}+f{{‘}^{2}}-c’}}$$.

System of Circles Key Points:

The equation of the common chord of the intersecting circles S = 0 is S – S’ = 0.

The equation of the common tangent of the touching circles S = 0 and S’ = 0 is S – S’ = 0.

If the circle S = 0 and the line L = 0 are intersecting then the equation of the circle passing through the point of intersection of S = 0 and L = 0 is S + λ L = 0.

The equation of the circle passing through the point of intersection of S = 0 and S’ = 0 is S + λS’ = 0.

Example: Find ‘k’ if the following pair of circles are orthogonal x² + y² + 2by – k = 0, x² + y² + 2ax + 8 = 0.

Solution: Given circles are x² + y² + 2by – k = 0, x² + y² + 2ax + 8 = 0 from above equations

g₁ = 0; f₁ = b; c₁ = -k;

g₂ = a; f₁ = 0; c₁ = 8.

Since the circles are orthogonal,

2g₁g₂ + 2f₁f₂ = c₁ + c₂

2 (0) (a) + 2 (b) (0) = -k + 8

0 = -k + 8

k = 8.