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

(A) The family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. Substituting the gi

Vedclass · Accepted Answer

$x^2+y^2+10x+3y+5=0$

Vedclass · Answer

(A) The family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. Substituting the given equations: $(x^2+y^2+2x+3y+1) + \lambda(x^2+y^2+4x+3y+2) = 0$. Since the circle passes through the point $(-1, 1)$,we substitute $x = -1$ and $y = 1$: $((-1)^2 + (1)^2 + 2(-1) + 3(1) + 1) + \lambda((-1)^2 + (1)^2 + 4(-1) + 3(1) + 2) = 0$. $(1 + 1 - 2 + 3 + 1) + \lambda(1 + 1 - 4 + 3 + 2) = 0$. $4 + 3\lambda = 0 \Rightarrow \lambda = -\frac{4}{3}$. Substituting $\lambda = -\frac{4}{3}$ back into the family equation: $(x^2+y^2+2x+3y+1) - \frac{4}{3}(x^2+y^2+4x+3y+2) = 0$. $3(x^2+y^2+2x+3y+1) - 4(x^2+y^2+4x+3y+2) = 0$. $3x^2+3y^2+6x+9y+3 - 4x^2-4y^2-16x-12y-8 = 0$. $-x^2-y^2-10x-3y-5 = 0$. $x^2+y^2+10x+3y+5 = 0$.

The equation of the circle passing through the points of intersection of two circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ and the point $(-1,1)$ is

Similar Questions

If the equation of the circle which cuts each of the circles $x^2+y^2=4$,$x^2+y^2-6x-8y+10=0$,and $x^2+y^2+2x-4y-2=0$ at the extremities of a diameter of these circles is $x^2+y^2+2gx+2fy+c=0$,then $g+f+c=$

The equation of the circle which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$ is

The equation of a circle passing through the points of intersection of the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+4y-12=0$ and having radius $\sqrt{13}$ is

Find the minimum radius of the circle which is orthogonal to both the circles $x^2+y^2+4x+3=0$ and $x^2+y^2-12x+35=0$.

The radical centre of the three circles $x^2+y^2-1=0$,$x^2+y^2-8x+15=0$ and $x^2+y^2+10y+24=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module