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

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

Find the equation of a circle with radius $3$ that touches the circle $x^{2} + y^{2} - 4x - 6y - 12 = 0$ internally at the point $(-1, -1)$.

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

If the centre of a circle,which passes through the points of intersection of the circles $x^2 + y^2 - 6x + 2y + 4 = 0$ and $x^2 + y^2 + 2x - 4y - 6 = 0$,lies on the line $y = x$,then the equation of the circle is:

$\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S_1: x^2+y^2-2x+6y=0$,$S_2: x^2+y^2+2gx-2y+6=0$,and $S_3: x^2+y^2-12x+2fy+3=0$. If $S_2$ and $S_3$ intersect orthogonally,then $(g, f) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module