The equation of the circle which passes throu | vedclass

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

Question

(C) Let the equation of the family of circles passing through the intersection of $S_1 = x^2 + y^2 - 8x - 2y + 7 = 0$ and $S_2 = x^2 + y^2 - 4x + 10y

Vedclass · Accepted Answer

$x^2 + y^2 + 22y + 9 = 0$

Vedclass · Answer

(C) Let the equation of the family of circles passing through the intersection of $S_1 = x^2 + y^2 - 8x - 2y + 7 = 0$ and $S_2 = x^2 + y^2 - 4x + 10y + 8 = 0$ be $S_1 + \lambda S_2 = 0$.$(x^2 + y^2 - 8x - 2y + 7) + \lambda(x^2 + y^2 - 4x + 10y + 8) = 0$$(1 + \lambda)x^2 + (1 + \lambda)y^2 - (8 + 4\lambda)x + (10\lambda - 2)y + (7 + 8\lambda) = 0$Dividing by $(1 + \lambda)$:$x^2 + y^2 - \frac{8 + 4\lambda}{1 + \lambda}x + \frac{10\lambda - 2}{1 + \lambda}y + \frac{7 + 8\lambda}{1 + \lambda} = 0$The centre of this circle is $(\frac{4 + 2\lambda}{1 + \lambda}, -\frac{5\lambda - 1}{1 + \lambda})$.Since the centre lies on the $y$-axis,the $x$-coordinate of the centre must be $0$.$\frac{4 + 2\lambda}{1 + \lambda} = 0 \implies 4 + 2\lambda = 0 \implies \lambda = -2$.Substituting $\lambda = -2$ into the family equation:$(x^2 + y^2 - 8x - 2y + 7) - 2(x^2 + y^2 - 4x + 10y + 8) = 0$$x^2 + y^2 - 8x - 2y + 7 - 2x^2 - 2y^2 + 8x - 20y - 16 = 0$$-x^2 - y^2 - 22y - 9 = 0$$x^2 + y^2 + 22y + 9 = 0$.

The equation of the circle which passes through the point of intersection of circles $x^2 + y^2 - 8x - 2y + 7 = 0$ and $x^2 + y^2 - 4x + 10y + 8 = 0$ and has its centre on the $y$-axis is:

Similar Questions

If $P$ and $Q$ are the points of intersection of the circles $x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $x^2 + y^2 + 2x + 2y - p^2 = 0$,then for what value of $p$ does a circle pass through $P, Q$ and $(1, 1)$?

The equation of the circle passing through the points of intersection of the circles $x^2 + y^2 - 6x + 8 = 0$ and $x^2 + y^2 - 6 = 0$ and the point $(1, 1)$ is:

The condition for the coaxial system $x^2+y^2+2 \lambda x+c=0$,where $\lambda$ is a parameter and $c$ is a constant,to have distinct limiting points,is

Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 - 8x + y - 15 = 0$ and $x^2 + y^2 - 4x + 4y - 42 = 0$.

The radical axis of circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module