The number of common tangents that can be dra | vedclass

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

Question

(D) For the circle $x^2+y^2-6x=0$,the center $C_1 = (3, 0)$ and radius $r_1 = \sqrt{3^2+0^2-0} = 3$.For the circle $x^2+y^2+6x+2y+1=0$,the center $C_2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) For the circle $x^2+y^2-6x=0$,the center $C_1 = (3, 0)$ and radius $r_1 = \sqrt{3^2+0^2-0} = 3$.For the circle $x^2+y^2+6x+2y+1=0$,the center $C_2 = (-3, -1)$ and radius $r_2 = \sqrt{(-3)^2+(-1)^2-1} = \sqrt{9+1-1} = 3$.The distance between the centers $d = C_1C_2 = \sqrt{(3 - (-3))^2 + (0 - (-1))^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$.Since $r_1 + r_2 = 3 + 3 = 6$ and $\sqrt{36}  r_1 + r_2$.Because the distance between the centers is greater than the sum of the radii,the two circles lie outside each other without touching.Therefore,the number of common tangents that can be drawn is $4$.

The number of common tangents that can be drawn to the circles $x^2+y^2-6x=0$ and $x^2+y^2+6x+2y+1=0$ is .....

Similar Questions

If $x-4=0$ is the radical axis of two orthogonal circles,out of which one is $x^2+y^2=36$,then the centre of the other circle is

Let the set of all values of $r$,for which the circles $(x+1)^{2}+(y+4)^{2}=r^{2}$ and $x^{2}+y^{2}-4x-2y-4=0$ intersect at two distinct points,be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

The equation of the circle passing through the point $(1, 2)$ and through the points of intersection of $x^2 + y^2 - 4x - 6y - 21 = 0$ and $3x + 4y + 5 = 0$ is given by

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k$ is equal to

If the points of intersection of the ellipses $x^{2}+2y^{2}-6x-12y+23=0$ and $4x^{2}+2y^{2}-20x-12y+35=0$ lie on a circle of radius $r$ and centre $(a, b)$,then the value of $ab+18r^{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module