Name: Two Circles Touching Internally and Externally
Uploaded: 2019-07-19
Duration: 6 min 6 s
Description: In this video, we will learn the condition for two circles touching internally and externally using an interesting example. Come, let's learn and watch this...

Question

Find the equations of circles passing through

(1, - 1)

touching the lines

4 x + 3 y + 5 = 0

and

3 x - 4 y - 10 = 0

.

Accepted Answer

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

Since the circle passes through $(1, - 1)$ .

So, the point satisfies its equation.

$\Rightarrow (1)^{2} + (- 1)^{2} + 2 g (1) + 2 f (- 1) + c = 0$

$\Rightarrow 2 + 2 g - 2 f + c = 0 \to (i)$

Equation of the lines touching the circles are

$4 x + 3 y + 5 = 0 \to (i i)$

$(1, - 1)$

On substituting the point $(1, - 1)$ in LHS of line $(i i)$ , we have

$4 (1) + 3 (- 1) + 5 = 6 > 0$

We know that the centre of the circle and any point on the tangent of the circle always lies on the same side of the tangent line

Two given points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ will lie on the same side of the line $a x + b y + c = 0$ if $a x_{1} + b y_{1} + c$ and $a x_{2} + b y_{2} + c$ will have same signs.

Thus, on substituting the centre $(- g, - f)$ on line $(i i)$ , we get the positive value.

$∴ |- 4 g - 3 f + 5| = - 4 g - 3 f + 5 \to (i v)$

Similarly, on substituting the point $(1, - 1)$ in LHS of line $(i i i)$ , we have

$3 + 4 - 10 = - 3 < 0$

Thus, on substituting the coordinates of centre in the equation of line we get the negative value.

$∴ |- 3 g + 4 f - 10| = - (- 3 g + 4 f - 10)$

$= 3 g - 4 f + 10 ⟶ (v)$

Now, the perpendicular distance from the centre $(- g, - f)$ to the tangent line $=$ Radius of the circle

$\Rightarrow \frac{- 4 g - 3 f + 5}{5} = \frac{3 g - 4 f + 10}{5} = \sqrt{(- g - 1)^{2} + (- f + 1)^{2}} \to (v i)$

From first two equations, we get

$- 4 g - 3 f + 5 = 3 g - 4 f + 10$

$\Rightarrow - 7 g + f - 5 = 0$

$\Rightarrow f = 7 g + 5 \to (v i i)$

Consider,

$\frac{3 g - 4 (7 g + 5) + 10}{5} = \sqrt{{(- g - 1)}^{2} + {(- f + 1)}^{2}}$

On substituting the value of $f$ , we have

$\frac{3 g - 4 (7 g + 5) + 10}{5} = \sqrt{(- g - 1)^{2} + (- f + 1)^{2}}$

$\Rightarrow \frac{- 25 g - 10}{5} = \sqrt{(- g - 1)^{2} + (- f + 1)^{2}}$

$\Rightarrow (- 5 g - 2)^{2} = (- g - 1)^{2} + (- f + 1)^{2}$

$\Rightarrow (- 5 g - 2)^{2} = (- g - 1)^{2} + (- 7 g - 4)^{2}$

$\Rightarrow 25 g^{2} + 4 + 20 g = g^{2} + 2 g + 1 + 49 g^{2} + 16 + 56 g$

$\Rightarrow 25 g^{2} + 38 g + 13 = 0$

$\Rightarrow g = \frac{- 38 \pm \sqrt{1444 - 1300}}{50}$

$\Rightarrow g = \frac{- 38 + 12}{50}$ or $g = \frac{- 38 - 12}{50}$

$\Rightarrow g = - \frac{13}{25}$ or $g = - 1$

From $(v i i)$ , we have

When $g = - 1 \Rightarrow$ $f = - 2$ and when $g = - \frac{13}{25} \Rightarrow$ $f = \frac{34}{25}$

Substituting these values in $(i)$ , we get

$2 + 2 g - 2 f + c = 0$

$\Rightarrow 2 + 2 (- \frac{13}{25}) - 2 (\frac{34}{25}) + c = 0$

$\Rightarrow 50 - 26 - 68 + 25 c = 0$

$\Rightarrow c = \frac{44}{25}$

Thus, the values of $g, f,$ and $c$ are $\frac{- 13}{25}, \frac{34}{25}$ , and $\frac{44}{25}$ .

Similarly, on substituting, $g = - 1, f = - 2$ in $(i)$ , we have $c = - 4$

Hence, the required equation of the circles are

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

And,

$x^{2} + y^{2} - \frac{26}{25} x + \frac{68}{25} y + \frac{44}{25} = 0$

Find the equations of circles passing through $(1, - 1)$ touching the lines $4 x + 3 y + 5 = 0$ and $3 x - 4 y - 10 = 0$ .

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the equations of circles passing through 1,-1 touching the lines 4x+3y+5=0 and 3x-4y-10=0.

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the equations of circles passing through $(1, - 1)$ touching the lines $4 x + 3 y + 5 = 0$ and $3 x - 4 y - 10 = 0$ .