Name: Tangents to a Circle From an External Point
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: Do you know the equation of the tangents to a circle drawn from an external point? Let's find the equation involving the slope, which helps to find the equation of tangents.

Question

A circle of radius

2

units rolls on the outside of the circle

x^{2} + y^{2} + 4 x = 0

touching it externally. Find the locus of the centre of this outer circle. Also, find the equations of the common tangents of the two circles when the line joining the centres of the two circles make an angle of

60 °

with

x

-axis.

Accepted Answer

We have,

$x^{2} + y^{2} + 4 x = 0$

$\Rightarrow {(x + 2)}^{2} + y^{2} = 4$

Centre $\equiv (- 2, 0)$ and radius $= 2$ units.

$C P^{2} = 4^{2} \Rightarrow (h + 2)^{2} + k^{2} = 16$

Since the two circles touch externally there will be only three common tangents.

One will be perpendicular to line of centres and other two will be parallel to the line of centres as the circles are of equal radius. The centre $P$ is given by

$\frac{h +2}{\cos 60 °} = \frac{k - 0}{\sin 60 °} = 4$

$∴ P (0, 2 \sqrt{3})$

Since, $M$ is mid-point of $C P$ , so $M (- 1, \sqrt{3})$

Slope of $C P$ is $\tan 60 ° = \sqrt{3}$

$∴$ Slope of line perpendicular to $C P$ is $- \frac{1}{\sqrt{3}}$ .

Equation of perpendicular line passing through $M (- 1, \sqrt{3})$ is

$y - \sqrt{3} = - \frac{1}{\sqrt{3}} (x + 1)$

$\Rightarrow x + \sqrt{3} y - 2 = 0$

which is one common tangent.

The other two will be parallel to $C P$ .

$∴ \sqrt{3} x - y + k = 0$ .

Applying condition of tangency

$\frac{0 - 2 \sqrt{3} + k}{2} = \pm 2$

$\Rightarrow k = 2 \sqrt{3} \pm 4$

Other common tangents are

$\sqrt{3} x - y \pm 4 + 2 \sqrt{3} = 0$

Important Questions on Circle

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