The equation of the circle passing through th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The family of circles passing through the intersection of $S_1: x^2 + y^2 - 1 = 0$ and $S_2: x^2 + y^2 - 2x - 4y + 1 = 0$ is given by $S_2 + \lamb

Vedclass · Accepted Answer

$x^2 + y^2 - x - 2y = 0$

Vedclass · Answer

(C) The family of circles passing through the intersection of $S_1: x^2 + y^2 - 1 = 0$ and $S_2: x^2 + y^2 - 2x - 4y + 1 = 0$ is given by $S_2 + \lambda S_1 = 0$.$(x^2 + y^2 - 2x - 4y + 1) + \lambda(x^2 + y^2 - 1) = 0$$(1 + \lambda)x^2 + (1 + \lambda)y^2 - 2x - 4y + (1 - \lambda) = 0$Dividing by $(1 + \lambda)$,we get $x^2 + y^2 - \frac{2}{1 + \lambda}x - \frac{4}{1 + \lambda}y + \frac{1 - \lambda}{1 + \lambda} = 0$.The centre is $C = \left( \frac{1}{1 + \lambda}, \frac{2}{1 + \lambda} \right)$ and the radius $r = \sqrt{\left( \frac{1}{1 + \lambda} \right)^2 + \left( \frac{2}{1 + \lambda} \right)^2 - \frac{1 - \lambda}{1 + \lambda}} = \sqrt{\frac{1 + 4 - (1 - \lambda)(1 + \lambda)}{(1 + \lambda)^2}} = \sqrt{\frac{5 - (1 - \lambda^2)}{(1 + \lambda)^2}} = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$.Since the circle touches the line $x + 2y = 0$,the perpendicular distance from the centre to the line equals the radius:$\left| \frac{\frac{1}{1 + \lambda} + 2(\frac{2}{1 + \lambda})}{\sqrt{1^2 + 2^2}} \right| = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$$\left| \frac{5}{\sqrt{5}(1 + \lambda)} \right| = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$$\sqrt{5} = \sqrt{4 + \lambda^2}$ $\Rightarrow 5 = 4 + \lambda^2$ $\Rightarrow \lambda^2 = 1$ $\Rightarrow \lambda = \pm 1$.For $\lambda = -1$,the equation does not represent a circle. Thus,$\lambda = 1$.Substituting $\lambda = 1$ into the family equation: $2x^2 + 2y^2 - 2x - 4y = 0$,which simplifies to $x^2 + y^2 - x - 2y = 0$.

The equation of the circle passing through the points of intersection of $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$ is

Similar Questions

If the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$ touch externally,then $\lambda$ is equal to:

When do the two given circles $x^2 + y^2 + ax + by + c = 0$ and $x^2 + y^2 + dx + ey + f = 0$ intersect each other orthogonally?

The number of direct common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 8x - 8y + 7 = 0$ is:

The angle of intersection of the circles $x^2 + y^2 - x + y - 8 = 0$ and $x^2 + y^2 + 2x + 2y - 11 = 0$ is

$A$ circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally,then the equation of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module