The radius of the circle passing through the | vedclass

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

Question

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

Vedclass · Accepted Answer

$\sqrt{39}$

Vedclass · Answer

(C) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2+2x+4y+1=0$ and $S_2: x^2+y^2-2x-4y-4=0$ is given by $S_1 + \lambda S_2 = 0$. $(1+\lambda)x^2 + (1+\lambda)y^2 + (2-2\lambda)x + (4-4\lambda)y + (1-4\lambda) = 0$. Dividing by $(1+\lambda)$,we get $x^2+y^2 + \frac{2(1-\lambda)}{1+\lambda}x + \frac{4(1-\lambda)}{1+\lambda}y + \frac{1-4\lambda}{1+\lambda} = 0$. This circle intersects $x^2+y^2-6=0$ orthogonally. The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1 = \frac{1-\lambda}{1+\lambda}, f_1 = \frac{2(1-\lambda)}{1+\lambda}, c_1 = \frac{1-4\lambda}{1+\lambda}$ and $g_2=0, f_2=0, c_2=-6$. Substituting these,$2(\frac{1-\lambda}{1+\lambda})(0) + 2(\frac{2(1-\lambda)}{1+\lambda})(0) = \frac{1-4\lambda}{1+\lambda} - 6$. $0 = \frac{1-4\lambda - 6 - 6\lambda}{1+\lambda} \implies 1-4\lambda-6-6\lambda = 0 \implies -10\lambda = 5 \implies \lambda = -1/2$. Substituting $\lambda = -1/2$ into the circle equation: $(1-1/2)x^2 + (1-1/2)y^2 + (2+1)x + (4+2)y + (1+2) = 0$. $0.5x^2 + 0.5y^2 + 3x + 6y + 3 = 0 \implies x^2+y^2+6x+12y+6=0$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{3^2+6^2-6} = \sqrt{9+36-6} = \sqrt{39}$.

The radius of the circle passing through the points of intersection of the circles $x^2+y^2+2x+4y+1=0$ and $x^2+y^2-2x-4y-4=0$ and intersecting the circle $x^2+y^2=6$ orthogonally is

Similar Questions

The equation of the 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 cutting the circle $x^2+y^2-4x+4y+8=0$ orthogonally is

The coordinates of the point from where the tangents drawn to the circles ${x^2} + {y^2} = 1$,${x^2} + {y^2} + 8x + 15 = 0$,and ${x^2} + {y^2} + 10y + 24 = 0$ are of the same length are:

If $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p$,where $0 < p < a$,then $\lambda=$

If two circles touch at $(0,0)$ externally,then their radical axis is

Find the equation of the circle passing through the intersection of the circles $x^{2} + y^{2} - 8x - 2y + 7 = 0$ and $x^{2} + y^{2} - 4x + 10y + 8 = 0$,and having its center on the $y-$axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module