If the circles x^{2}+y^{2}=9 and x^{2}+y^{2}+ | vedclass

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

Question

(A) The centers and radii of the given circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ are $C_{1}(0,0), r_{1}=3$ and $C_{2}(-\alpha, -1),

Vedclass · Accepted Answer

$\pm \frac{4}{3}$

Vedclass · Answer

(A) The centers and radii of the given circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ are $C_{1}(0,0), r_{1}=3$ and $C_{2}(-\alpha, -1), r_{2}=\sqrt{(-\alpha)^{2}+(-1)^{2}-1} = \sqrt{\alpha^{2}} = |\alpha|$.Since the two circles touch each other internally,the distance between their centers must be equal to the difference of their radii:$C_{1}C_{2} = |r_{1} - r_{2}|$$\sqrt{(-\alpha - 0)^{2} + (-1 - 0)^{2}} = |3 - |\alpha||$$\sqrt{\alpha^{2} + 1} = |3 - |\alpha||$Squaring both sides:$\alpha^{2} + 1 = (3 - |\alpha|)^{2}$$\alpha^{2} + 1 = 9 - 6|\alpha| + \alpha^{2}$$6|\alpha| = 8$$|\alpha| = \frac{4}{3}$$\alpha = \pm \frac{4}{3}$

If the circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ touch each other internally,then $\alpha$ is equal to

Similar Questions

If the circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts the three circles $x^{2}+y^{2}-5=0$, $x^{2}+y^{2}-8x-6y+10=0$, and $x^{2}+y^{2}-4x+2y-2=0$ at the extremities of their diameters, then

The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point

The angle of intersection of the circles $x^2 + y^2 - x + y - 8 = 0$ and $x^2 + y^2 + 2x + 2y - 11 = 0$ is

If the length of the transverse common tangent to the circles $x^{2} + y^{2} = 1$ and $(x - h)^{2} + y^{2} = 1$ is $2\sqrt{3}$,find the value of $h$.

If the lengths of the tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the coordinates of point $P$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module