Question

If y = 2x is a chord of the circle $x^2+y^2-10x=0$, find the equation of a circle with this chord as diameter.

Or

Find the equation of ellipse whose foci are (2, 3) and (- 2, 3) and whose semi-minor axis is $\sqrt{5}$

Answer 1

The points of intersection of the given chord and the given circle are obtained by solving y = 2x and $x^2+y^2-10x=0$ simultaneously.

Putting y = 2x in $x^2+y^2$ -10x = 0, we get

$x^2+4x^2-10x=0\Rightarrow 5x^2-10x=0$

$\Rightarrow 5x(x-2)=0\Rightarrow x=0,2$

Putting x = 0 and x = 2, respectively in y = 2x, we get y = 0 and y = 4.

Thus, the coordinates of the point of intersection of the given chord and given circle is A(0, 0) and B(2, 4).

The equation of circle with chord AB as diameter is

(x - 0)(x - 2) + (y - 0)(y - 4) = 0

[$\because$ equation of circle having end points of a diameter $(x_1,y_1)$ and $(x_2,y_2)$ is $(x-x_1)$ $(x-x_2)$ - $(y-y_1)$$(y-y_2)$= 0]

$\therefore x^2+y^2-2x-4y=0$

Or

Let S and S' be two foci of the required ellipse.

Then, the coordinates of S and S' are (2, 3) and (- 2, 3), respectively.

Now, $S{S}^{\prime }=\sqrt{(2+2{)}^2+(3-3{)}^2}=4$

Let 2a and 2b be the length of the axes of the ellipse and e be its eccentricity.

Then, SS' =2c

$\Rightarrow$ 2c = 4 $\Rightarrow$ c = 2

Now, $b^2=a^2-C^2$

$\Rightarrow (\sqrt{5}{)}^2=a^2-(2{)}^2\Rightarrow a^2=5+4=9$

[$\because$ semi-minor axis, $b=\sqrt{5}$]

$\Rightarrow$ a = 3

Let P (x, y) be any point on the ellipse. Then, by definition of ellipse

SP + S' P = 2a

$\Rightarrow \sqrt{(x-2{)}^2+(y-3{)}^2}+\sqrt{(x+2{)}^2+(y-3{)}^2}=2\times 3$

$\Rightarrow \sqrt{(x-2{)}^2+(y-3{)}^2}=6-\sqrt{(x+2{)}^2+(y-3{)}^2}$

On squaring both sides, we get

$(x-2{)}^2+(y-3{)}^2=36+(x+2{)}^2+(y-3{)}^2-12\sqrt{(x+2{)}^2+(y-3{)}^2}$

$\Rightarrow x^2+4-4x+y^2+9-6y=36+x^2+4+4x+y^2+9-6y-12\sqrt{(x+2{)}^2+(y-3{)}^2}$

$\Rightarrow -8x-36=-12\sqrt{(x+2{)}^2+(y-3{)}^2}$

$\Rightarrow (2x+9)=3\sqrt{(x+2{)}^2+(y-3{)}^2}$

[divide both sides by - 4]

$\Rightarrow (2x+9{)}^2=9[(x+2{)}^2+(y-3{)}^2]$

[on squaring both sides]

$\Rightarrow 4x^2+81+36x=9[x^2+4+4x+y^2+9-6y]$

$\therefore 5x^2+9y^2-54y+36=0$

which is the required equation of the ellipse.