Name: Definition and Equation of a Circle
Uploaded: 2019-07-19
Duration: 2 min 25 s
Description: Coins are circular in shape. Even many cricket stadiums are circular, and so is pizza. How are these circles made, and how can we make them perfectly circula...

Question

The line

2 x - y + 6 = 0

meets the circle

x^{2} + y^{2} + 10 x + 9 = 0

at

A

and

B

. Find the equation of circle on

A B

as diameter.

Accepted Answer

Given that, the line $2 x - y + 6 = 0 \Rightarrow y = 2 x + 6$ ....(i) meets the circle $x^{2} + y^{2} + 10 x + 9 = 0$ at two points $A$ and $B$ .

To find the points of intersection of line with a circle, substitute equation (i) in the given circle equation.

$\Rightarrow x^{2} + (2 x + 6)^{2} + 10 x + 9 = 0$

$\Rightarrow x^{2} + 4 x^{2} + 24 x + 36 + 10 x + 9 = 0$

$\Rightarrow 5 x^{2} + 34 x + 45 = 0$

$\Rightarrow 5 x^{2} + 25 x + 9 x + 45 = 0$

$\Rightarrow 5 x (x + 5) + 9 (x + 5) = 0$

$\Rightarrow (5 x + 9) (x + 5) = 0$

$\Rightarrow x = \frac{- 9}{5}, - 5$

Substitute the values of $x$ in equation (i), we get

$x = \frac{- 9}{5} \Rightarrow y = 2 (- \frac{9}{5}) + 6 = \frac{12}{5} \Rightarrow A (- \frac{9}{5}, \frac{12}{5})$

$x = - 5 \Rightarrow y = 2 (- 5) + 6 = - 4 \Rightarrow B (- 5, - 4)$

Equation of circle with $A B$ as diameter is

$(x + \frac{9}{5}) (x + 5) + (y - \frac{12}{5}) (y + 4) = 0$

$x^{2} + y^{2} + \frac{34 x}{5} + \frac{8 y}{5} - \frac{3}{5} = 0$

Therefore, the required equation of a circle is $5 x^{2} + 5 y^{2} + 34 x + 8 y - 3 = 0$

The line $2 x - y + 6 = 0$ meets the circle $x^{2} + y^{2} + 10 x + 9 = 0$ at $A$ and $B$ . Find the equation of circle on $A B$ as diameter.

Important Questions on Circle

Learn and Prepare for any exam you want !

The line 2x−y+6=0 meets the circle x2+y2 +10x+9=0 at A and B. Find the equation of circle on AB as diameter.

Important Questions on Circle

Learn and Prepare for any exam you want !

The line $2 x - y + 6 = 0$ meets the circle $x^{2} + y^{2} + 10 x + 9 = 0$ at $A$ and $B$ . Find the equation of circle on $A B$ as diameter.