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 length of the chord intercepted by the circle

x^{2} + y^{2} - 8 x - 2 y - 8 = 0

on the line

x + y + 1 = 0

.

Accepted Answer

Given equation of the circle is $x^{2} + y^{2} - 8 x - 2 y - 8 = 0 \to (i)$ and equation of the line is $x + y + 1 = 0$ .

We know that the general equation of the circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \to (i i)$

Comparing equations $(i)$ and $(i i)$ , we have

$g = - 4, f = - 1$ , and $c = - 8$

Now, centre of the given circle $= (- g, - f) =$ $(4, 1)$

Radius of the circle $= \sqrt{g^{2} + f^{2} - c}$

$= \sqrt{{(- 4)}^{2} + {(- 1)}^{2} + 8}$

$= \sqrt{16 + 1 + 8} = \sqrt{25} = 5$

Consider,

Now, let the line $x + y + 1 = 0$ intersects the circle at $A$ and $B$ . Then, we need to find the length of the chord $A B$ .

The Perpendicular distance from the centre $C (4, 1)$ to the line $x + y + 1 = 0$ is given by

$d = \frac{|4 + 1 + 1|}{\sqrt{(1)^{2} + (1)^{2}}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$

Now, from the figure, $B P^{2} = B C^{2} - P C^{2}$

$= r^{2} - d^{2}$

$= {(5)}^{2} - {(3 \sqrt{2})}^{2}$

$= 25 - 18$

$\Rightarrow B P^{2} = 7$

$\Rightarrow B P = \sqrt{7}$ units

We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.

$∴ A B = 2 B P$

$= 2 \sqrt{7}$ units

Hence, the required length of the chord is $2 \sqrt{7}$ units.

Find the length of the chord intercepted by the circle $x^{2} + y^{2} - 8 x - 2 y - 8 = 0$ on the line $x + y + 1 = 0$ .

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the length of the chord intercepted by the circle x2+y2-8x-2y-8=0 on the line x+y+1=0.

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the length of the chord intercepted by the circle $x^{2} + y^{2} - 8 x - 2 y - 8 = 0$ on the line $x + y + 1 = 0$ .