# If x-y=2 is the equation of a chord of the circle x^2+y^2+2y=0.Find the equation of the circle of which this chord is a diameter.?

Nov 9, 2017

Given that

• the equation of the chord $\text{ } \textcolor{m a \ge n t a}{x - y = 2. \ldots . \left[1\right]}$

• the equation of the circle $\text{ } \textcolor{b l u e}{{x}^{2} + {y}^{2} + 2 y = 0. \ldots . . \left[2\right]}$

From these two equations we get

${\left(y + 2\right)}^{2} + {y}^{2} + 2 y = 0$

$\implies 2 {y}^{2} + 6 y + 4 = 0$

$\implies 2 {y}^{2} + 6 y + 4 = 0$

$\implies {y}^{2} + 3 y + 2 = 0$

$\implies {y}^{2} + 2 y + y + 2 = 0$

$\implies y \left(y + 2\right) + 1 \left(y + 2\right) = 0$

$\implies \left(y + 2\right) \left(y + 1\right) = 0$

So $y = - 1 \mathmr{and} - 2$

Inserting in [1] we get

$x = 1 \mathmr{and} 0$ respectively

So the coordinates of points intersections of the chord with the circles are $\left(1 , - 1\right) \mathmr{and} \left(0 , - 2\right)$
The line segment of the chord is also the diameter of a circle.

So equation of this circle having diameter $x - y = 2$ will be

$\frac{y + 1}{x - 1} \times \frac{y + 2}{x - 0} = - 1$

$\implies \left(y + 1\right) \times \left(y + 2\right) = - x \left(x - 1\right)$

$\textcolor{g r e e n}{\implies {x}^{2} + {y}^{2} - x + 3 y + 2 = 0. \ldots . \left[3\right]}$

This is the required equation of the circle

Nov 9, 2017

${x}^{2} + {y}^{2} - x + 3 y + 2 = 0 ,$

#### Explanation:

We will solve this Problem using the following Result R :

R : The Equation (eqn.) of a Circle that passes through

the Points (pt.) of Intersection of a Circle S & Line L

$S : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0 \mathmr{and} L : l x + m y + n = 0 ,$ is

$S + \lambda L : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c + \lambda \left(l x + m y + n\right) = 0 ,$

where, $\lambda \in \mathbb{R} .$

Let, $S : {x}^{2} + {y}^{2} + 2 y = 0 , \mathmr{and} L : x - y - 2 = 0.$

Note that, the Reqd. Circle, say $S ' ,$ passes through the pts. of

$S \cap L .$

Applying R, we may suppose that, $S ' : S + \lambda L = 0 , i . e . ,$

$S ' : {x}^{2} + {y}^{2} + 2 y + \lambda \left(x - y - 2\right) = 0.$

$\therefore S ' : {x}^{2} + \lambda x + {y}^{2} + \left(2 - \lambda\right) y = 2 \lambda .$

$\therefore S ' : {x}^{2} + \lambda x + {\lambda}^{2} / 4 + {y}^{2} + \left(2 - \lambda\right) y + {\left(2 - \lambda\right)}^{2} / 4 = 2 l a m \mathrm{da} + {\lambda}^{2} / 4 + {\left(2 - \lambda\right)}^{2} / 4 , \mathmr{and} ,$

$S ' : {\left(x + \frac{\lambda}{2}\right)}^{2} + {\left(y + \frac{2 - \lambda}{2}\right)}^{2} = 2 l a m \mathrm{da} + {\lambda}^{2} / 4 + {\left(2 - \lambda\right)}^{2} / 4.$

This shows that the Centre $C$ of $S '$ is $C \left(- \frac{\lambda}{2} , - \frac{2 - \lambda}{2}\right) .$

Now, given that $L$ is a diameter of $S ' \Rightarrow C \in L .$

$\Rightarrow - \frac{\lambda}{2} - \left(- \frac{2 - \lambda}{2}\right) = 2.$

$\therefore - \frac{\lambda}{2} + \frac{2 - \lambda}{2} = 2 , \mathmr{and} , - \lambda = 2 - 1 = 1 , i . e . , \lambda = - 1.$

Therefore, $S ' : {x}^{2} + {y}^{2} + 2 y - 1 \left(x - y - 2\right) = 0 , i . e . ,$

$S ' : {x}^{2} + {y}^{2} - x + 3 y + 2 = 0 ,$ is the reqd. eqn. of the Circle, as

already derived by Respected dk_ch Sir!

Enjoy Maths.!