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

(D) The equation of any circle passing through the intersection of the circle $S = x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $L = 3x + 4y + 5 = 0$ is

Vedclass · Accepted Answer

$x^2 + y^2 + 2x + 2y - 11 = 0$

Vedclass · Answer

(D) The equation of any circle passing through the intersection of the circle $S = x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $L = 3x + 4y + 5 = 0$ is given by $S + \lambda L = 0$.$(x^2 + y^2 - 4x - 6y - 21) + \lambda(3x + 4y + 5) = 0$Since this circle passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ into the equation:$(1^2 + 2^2 - 4(1) - 6(2) - 21) + \lambda(3(1) + 4(2) + 5) = 0$$(1 + 4 - 4 - 12 - 21) + \lambda(3 + 8 + 5) = 0$$-32 + 16\lambda = 0$$16\lambda = 32 \Rightarrow \lambda = 2$Substituting $\lambda = 2$ back into the family of circles equation:$(x^2 + y^2 - 4x - 6y - 21) + 2(3x + 4y + 5) = 0$$x^2 + y^2 - 4x - 6y - 21 + 6x + 8y + 10 = 0$$x^2 + y^2 + 2x + 2y - 11 = 0$Thus,the required equation is $x^2 + y^2 + 2x + 2y - 11 = 0$.

The equation of the circle passing through the point $(1, 2)$ and through the points of intersection of $x^2 + y^2 - 4x - 6y - 21 = 0$ and $3x + 4y + 5 = 0$ is given by

Similar Questions

If the radical centre of the circles $x^2+y^2-8x-2y+8=0$,$x^2+y^2+6x+8y-24=0$,and $x^2+y^2-2x+2y+2=0$ is $(a, b)$,then $a+b=$

If the circle $x^2+y^2-6x-12y+1=0$ cuts another circle $C$ orthogonally and the centre of the circle $C$ is $(-4, 2)$,then its radius 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

The radical axis of the co-axial system of circles with limiting points $(1, 2)$ and $(-2, 1)$ is

Let the latus rectum of the parabola $y^{2} = 4x$ be the common chord to the circles $C_{1}$ and $C_{2}$,each of them having radius $2\sqrt{5}$. Then,the distance between the centres of the circles $C_{1}$ and $C_{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module