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

Let the mirror image of the point

A (5, 6)

with respect to the line

2 x + 3 y = 15

be

B

. Find the equation of the circle described on

A B

as diameter.

A C

in any chord of the circle meeting the

x

-axis at

D

such that

A D = 10 D C

. How many such chords are there?

Accepted Answer

Given:

$L : 2 x + 3 y = 15 . . . (1)$

Slope of line $x$ is $D$

Slope of line $A B$ perpendicular to $L$ is $m_{2} = \frac{3}{2}$

$∴$ Equation of line $A B$

$y - y_{1} = m_{2} (x - x_{1})$

$\Rightarrow (y - 6) = \frac{3}{2} (x - 5)$

$\Rightarrow 3 x - 2 y = 3 . . . (2)$

Solving $(1) & (2)$ , we get

$x = 3$

$∴ y = \frac{15 - 2 x}{3} = \frac{15 - (3 \times 2)}{3} = 3$

$P (3, 3)$

Let image point is $B (a, b)$ , so

$\frac{5 + a}{2} = 3 \Rightarrow a = 1$

$\frac{6 + b}{2} = 3 \Rightarrow B = 0$

$[∵ A P = P B]$

$∴ (a, b) \equiv (1, 0)$

Now, equation of circle described on $A B$ as diameter

$(x - a) (x - 5) + (y - b) (y - 6) = 0$

$\Rightarrow (x - 1) (x - 5) + y (y - 6) = 0$

$\Rightarrow x^{2} + y^{2} - 6 x - 6 y + 5 = 0$

$\Rightarrow {(x - 3)}^{2} + {(y - 3)}^{2} = 13 . . . (3)$

Let, $D$ be the point on $X$ -axis having coordinate $(d, 0)$ and $\frac{A D}{D C} = 10$ (given), $C (h, k)$

By section formula,

$d = \frac{10 h + 5}{10 + 1}$ and $0 = \frac{10 k + 6}{10 + 1}$

$\Rightarrow k = - \frac{6}{10} = \frac{- 3}{5}$

$∴ (h, k) \equiv (h, - \frac{3}{5})$

Now, $C (h, k)$ lies on $(3)$

$\Rightarrow h^{2} + {(- \frac{3}{5})}^{2} - 6 h - 6 (- \frac{3}{5}) + 5 = 0$

$\Rightarrow h^{2} + \frac{9}{25} - 6 h + \frac{18}{5} + 5 = 0$

$\Rightarrow h^{2} - 6 h + \frac{9 + 90 + 125}{25} = 0$

$\Rightarrow 25 h^{2} - 150 h + 224 = 0$

$∴ h = \frac{- (- 150) \pm \sqrt{(- 150)^{2} - 4 (25) (224)}}{2 \times 25} = \frac{150 \pm 10}{50}$

$\Rightarrow h = \frac{14}{5}, \frac{16}{5}$

$∴$ Two possible points $C (\frac{16}{5}, \frac{- 3}{5})$ and $C (\frac{14}{5}, \frac{- 3}{5})$ can be obtained Such that $A D = 10 D C_{1}$ and $A D = 10 D C_{2}$

$∴$ Two such Chords are possible.

Let the mirror image of the point $A (5, 6)$ with respect to the line $2 x + 3 y = 15$ be $B$ . Find the equation of the circle described on $A B$ as diameter. $A C$ in any chord of the circle meeting the $x$ -axis at $D$ such that $A D = 10 D C$ . How many such chords are there?

Important Questions on Circle

Learn and Prepare for any exam you want !

Let the mirror image of the point A5,6 with respect to the line 2x+3y=15 be B. Find the equation of the circle described on AB as diameter. AC in any chord of the circle meeting the x-axis at D such that AD=10DC. How many such chords are there?

Important Questions on Circle

Learn and Prepare for any exam you want !

Let the mirror image of the point $A (5, 6)$ with respect to the line $2 x + 3 y = 15$ be $B$ . Find the equation of the circle described on $A B$ as diameter. $A C$ in any chord of the circle meeting the $x$ -axis at $D$ such that $A D = 10 D C$ . How many such chords are there?