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) The given circles are $x^{2}+y^{2}-y=0$ and $x^{2}+y^{2}+y=0$. For the first circle $x^{2}+y^{2}-y=0$,the center is $C_{1}(0, 1/2)$ and radius $r_

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The given circles are $x^{2}+y^{2}-y=0$ and $x^{2}+y^{2}+y=0$. For the first circle $x^{2}+y^{2}-y=0$,the center is $C_{1}(0, 1/2)$ and radius $r_{1} = \sqrt{0^{2} + (1/2)^{2} - 0} = 1/2$. For the second circle $x^{2}+y^{2}+y=0$,the center is $C_{2}(0, -1/2)$ and radius $r_{2} = \sqrt{0^{2} + (-1/2)^{2} - 0} = 1/2$. The distance between the centers is $C_{1}C_{2} = \sqrt{(0-0)^{2} + (1/2 - (-1/2))^{2}} = \sqrt{1^{2}} = 1$. Since $r_{1} + r_{2} = 1/2 + 1/2 = 1$,we have $C_{1}C_{2} = r_{1} + r_{2}$. This condition implies that the two circles touch each other externally at a single point. When two circles touch each other externally,they have exactly $3$ common tangents (two external and one internal).

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

Similar Questions

The circle $x^2 + y^2 - 4x - 4y + 4 = 0$ is...

If $x^2 + y^2 + 2gx + 2fy + c = 0$ is the equation of the smallest circle passing through $(1, 2)$ and touching the line $x + y - 7 = 0$,then the value of $(g + 2f + 3c)$ is -

The radius of a circle whose center is $(2, 1)$ and one of its chords is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is:

If the point $(\lambda, 1+\lambda)$ lies inside the circle $x^2+y^2=1$,then

Suppose a circle passes through $(0, a)$ and $(b, h)$ having its centre at $(c, 0)$. Then the value of $c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module