If left(0, frac{3}{4}right) is the radical ce | vedclass

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

Question

(C) The radical axis of $S$ and $S^{\prime}$ is given by $S-S^{\prime}=0$: $\Rightarrow (x^2+y^2+\alpha x+6y) - (x^2+y^2+2\alpha x+\alpha y+6) = 0$ $\

Vedclass · Accepted Answer

$\frac{\sqrt{65}}{4}$

Vedclass · Answer

(C) The radical axis of $S$ and $S^{\prime}$ is given by $S-S^{\prime}=0$: $\Rightarrow (x^2+y^2+\alpha x+6y) - (x^2+y^2+2\alpha x+\alpha y+6) = 0$ $\Rightarrow -\alpha x + (6-\alpha)y - 6 = 0$ Since $\left(0, \frac{3}{4}\right)$ lies on this axis: $-\alpha(0) + (6-\alpha)(\frac{3}{4}) - 6 = 0$ $\Rightarrow \frac{18-3\alpha}{4} = 6$ $\Rightarrow 18-3\alpha = 24$ $\Rightarrow -3\alpha = 6$ $\Rightarrow \alpha = -2$. Now,the circle $S^{\prime}$ is $x^2+y^2+2(-2)x+(-2)y+6 = 0$,which is $x^2+y^2-4x-2y+6 = 0$. The centre of $S^{\prime}$ is $C = (-g, -f) = (2, 1)$. The radical centre is $P = \left(0, \frac{3}{4}\right)$. The distance $PC = \sqrt{(2-0)^2 + (1-\frac{3}{4})^2} = \sqrt{4 + (\frac{1}{4})^2} = \sqrt{4 + \frac{1}{16}} = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4}$.

Similar Questions

If the acute angle between the circles $S \equiv x^2+y^2+2kx+4y-3=0$ and $S' \equiv x^2+y^2-4x+2ky+9=0$ is $\cos^{-1}(\frac{3}{8})$ and the centre of $S'=0$ lies in the first quadrant,then the radical axis of $S=0$ and $S'=0$ is

If one of the diameters of the curve $x^{2}+y^{2}-4x-6y+9=0$ is a chord of a circle with centre $(1,1)$, the radius of this circle is

$A$ circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1}$. Then the diameter of $C$ is:

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$,then

If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module