# Point of Contact – Circles

## Point of Contact – Circles

1. The condition that the straight-line lx + my + n = 0 may touch the circle x² + y² = a² is n² = a² (l² + m²) and the point of contact is $$\left( \frac{-{{a}^{2}}l}{n},\,\frac{-{{a}^{2}}m}{n} \right)$$. Proof: Given line is lx + my + n = 0 … (1)

Given circle is x² + y² = a² … (2)

Centre O (0, 0) and radius = a

Line (1) touches circle (2)

⟺ the perpendicular distance from O to line (1) is $$a=\left| \frac{l(0)+m(0)+n}{\sqrt{{{l}^{2}}+{{m}^{2}}}} \right|$$ $$\left( \because \,\,d=\left| \frac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right| \right)$$,

⟺ $$\left| \frac{n}{\sqrt{{{l}^{2}}+{{m}^{2}}}} \right|=a$$,

n² = a² (l² + m²)

Let P (x₁, y₁) be the point of contact.

The equation of the tangent to the circle (1) at P is xx₁ + yy₁ – a² = 0 … (3)

Now (1) and (3) represent the same line

∴ $$\frac{{{x}_{1}}}{l}=\frac{{{y}_{1}}}{m}=-\frac{{{a}^{2}}}{n}$$,

$$\Rightarrow {{x}_{1}}=-\frac{{{a}^{2}}l}{n},\,\,{{y}_{1}}=-\frac{{{a}^{2}}m}{n}$$ ,

Point of contact$$=\left( \frac{-{{a}^{2}}l}{n},\,\frac{-{{a}^{2}}m}{n} \right)$$.

2. The condition for the straight line lx + my + n = – may be a tangent to the circle x² + y² + 2gx + 2fy + c = 0 is (g² + f² – c) (l² + m²) = (lg + mf -n)². Proof: The given line is lx + my + n = 0 … (1)

The given circle is x² + y² + 2gx + 2fy + c = 0 … (2)

Centre C = (-g, -f),

radius r = √g² + f² – c

Line (1) is a tangent to the circle (2)

⟺ The perpendicular distance from the center C to the line (1) is equal to the radius r

∵ $$\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\left| \frac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$$,

⟺ $$\left| \frac{-\lg -mf+n}{\sqrt{{{l}^{2}}+{{m}^{2}}}} \right|=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$$ (∵ The perpendicular distance equal to radius of the circle).

⟺ (lg + mf – n)² = (g² + f² – c) (l² + m²)

The condition for the straight-line y = mx + c to touch the circle x² + y² = a² is c² = a² (l + m²).

3. If the straight-line y = mx + c touches the circle x² + y² = a², then their point of contact is $$\left( -\frac{{{a}^{2}}m}{c},\,\frac{{{a}^{2}}}{c} \right)$$.

Proof: Let P (x₁, y₁) be the point of contact

Given line is y = mx + c

⟺ mx – y + c = 0 … (1)

The equation of tangent to the circle x² + y² = a² at P is xx₁ + yy₁ – a² = 0 … (2)

Now (1) and (2) represent the same line.

$$\frac{{{x}_{1}}}{m}=\frac{{{y}_{1}}}{-1}=\frac{-{{a}^{2}}}{c}$$,

⇒ $${{x}_{1}}=\frac{-{{a}^{2}}m}{c},\,\,{{y}_{1}}=\frac{{{a}^{2}}}{c}$$,

∴ Point of contact$$=\left( \frac{-{{a}^{2}}m}{c},\,\frac{{{a}^{2}}}{c} \right)$$.