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

Find the equations of the circles passing through

(- 4, 3)

and touching the lines

x + y = 2

and

x - y = 2

.

Accepted Answer

Let the equation of the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0 . . . (i)$

Where centre $C (- g, - f)$ and radius $= \sqrt{g^{2} + f^{2} - c}$

$∵$ It passes through $(- 4, 3)$

$∴ 16 + 9 - 8 g + 6 f + c = 0$

$\Rightarrow 25 - 8 g + 6 f + c = 0 . . . (i i)$

Since, circle touches the line $x + y - 2 = 0$ and $x - y - 2 = 0$ .

Therefore, perpendicular distance of both the lines from centre of circle is equal to radius of circle.

$∴ |\frac{- g - f - 2}{\sqrt{2}}| = |\frac{- g + f - 2}{\sqrt{2}}| = \sqrt{g^{2} + f^{2} - c} . . . (i i i)$

Now equating first two terms, we get

$|\frac{- g - f - 2}{\sqrt{2}}| = |\frac{- g + f - 2}{\sqrt{2}}|$

$\Rightarrow - g - f - 2 = \pm (- g + f - 2)$

$\Rightarrow - g - f - 2 = - g + f - 2$ or $- g - f - 2 = g - f + 2$

$\Rightarrow f = 0$ or $g = - 2$

Case I . When $f = 0$

From Eq. (iii), we get

$|\frac{- g - 2}{\sqrt{2}}| = \sqrt{g^{2} - c}$

$\Rightarrow (g + 2)^{2} = 2 (g^{2} - c)$

$\Rightarrow g^{2} - 4 g - 4 - 2 c = 0 . . . (i v)$

On putting $f = 0$ in Eq. (ii), we get

$25 - 8 g + c = 0 . . . (v)$

From (iv) and (v), on eliminating $c$ , we get

$g^{2} - 20 g + 46 = 0$

$\Rightarrow g = \frac{20 \pm \sqrt{400 - 4.1.46}}{2.1}$

$\Rightarrow g = 10 \pm 3 \sqrt{6}$

Putting value of $g$ in (v), we get $c = 55 \pm 24 \sqrt{6}$

On substituting the values of $g, f$ and $c$ in Eq. (i), we get equation of circle

$x^{2} + y^{2} + 2 (10 \pm 3 \sqrt{6}) x + (55 \pm 24 \sqrt{6}) = 0$

Case II :When $g = - 2$

From Eq. (iii), we get

$|\frac{f}{\sqrt{2}}| = \sqrt{4 + f^{2} - c}$

$\Rightarrow f^{2} = 2 (4 + f^{2} - c)$

$\Rightarrow f^{2} - 2 c + 8 = 0 . . . (v i)$

On putting $g = - 2$ in Eq. (ii), we get

$25 + 16 + 6 f + c = 0$

$c = - 6 f - 41 . . . (v i i)$

On substituting $c$ from equation (vii) in (vi), we get

$f^{2} + 12 f + 90 = 0$ :

Discriminant of this equation $D = 12^{2} - 4.1.90 < 0$

$\Rightarrow$ Equation does not have real values of $f$

Thus, there is no circle in this case.
Hence, the required equations of the circles are
$x^{2} + y^{2} + 2 (10 \pm 3 \sqrt{6}) x + (55 \pm 24 \sqrt{6}) = 0$

Find the equations of the circles passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the equations of the circles passing through (-4,3) and touching the lines x+y=2 and x-y=2.

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the equations of the circles passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .