The number of common tangents to the circles | vedclass

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

Question

(B) For the first circle $x^2+y^2+4x=0$,the center is $C_1 = (-2, 0)$ and the radius is $r_1 = \sqrt{(-2)^2 + 0^2 - 0} = 2$.For the second circle $x^2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) For the first circle $x^2+y^2+4x=0$,the center is $C_1 = (-2, 0)$ and the radius is $r_1 = \sqrt{(-2)^2 + 0^2 - 0} = 2$.For the second circle $x^2+y^2-2x=0$,the center is $C_2 = (1, 0)$ and the radius is $r_2 = \sqrt{(1)^2 + 0^2 - 0} = 1$.The distance between the centers is $C_1C_2 = \sqrt{(1 - (-2))^2 + (0 - 0)^2} = \sqrt{3^2} = 3$.Since $r_1 + r_2 = 2 + 1 = 3$,we have $C_1C_2 = r_1 + r_2$.This condition implies that the two circles touch each other externally.When two circles touch each other externally,the number of common tangents is $3$ (two direct common tangents and one transverse common tangent).

The number of common tangents to the circles $x^2+y^2+4x=0$ and $x^2+y^2-2x=0$ is

Similar Questions

$A$ circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$

If $y + 3x = 0$ is the equation of a chord of the circle $x^2 + y^2 - 30x = 0$,then the equation of the circle with this chord as diameter is:

The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y-7=0$ is

The ratio of the areas of the greatest and the smallest circles touching $(x \pm 1)^2 + (y \pm 1)^2 = 1$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module