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

(D) For the circle $S_1: x^2+y^2-2x-6y+9=0$,the centre $C_1$ is $(1, 3)$ and the radius $r_1 = \sqrt{1^2+3^2-9} = \sqrt{1} = 1$.For the circle $S_2: x

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) For the circle $S_1: x^2+y^2-2x-6y+9=0$,the centre $C_1$ is $(1, 3)$ and the radius $r_1 = \sqrt{1^2+3^2-9} = \sqrt{1} = 1$.For the circle $S_2: x^2+y^2+6x-2y+1=0$,the centre $C_2$ is $(-3, 1)$ and the radius $r_2 = \sqrt{(-3)^2+1^2-1} = \sqrt{9} = 3$.The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-3))^2 + (3 - 1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.Since $\sqrt{20} \approx 4.47$ and $r_1 + r_2 = 1 + 3 = 4$,we observe that $d > r_1 + r_2$.Because the distance between the centres is greater than the sum of the radii,the two circles do not intersect and lie outside each other.Therefore,the number of common tangents is $4$.

The number of common tangents to the circles $x^2+y^2-2x-6y+9=0$ and $x^2+y^2+6x-2y+1=0$ is

Similar Questions

If the ratio of the lengths of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ is $4 : 3$,then $k = \dots$

The equation of the chord of the circle $x^2 + y^2 = a^2$ having $(x_1, y_1)$ as its mid-point is

Find the circumcenter of the triangle formed by the points $(a \cos \alpha, a \sin \alpha)$,$(a \cos \beta, a \sin \beta)$,and $(a \cos \gamma, a \sin \gamma)$.

The area (in sq. units) of the region bounded by the circle $x^2+y^2=64$, positive $x$-axis and the line $y=\sqrt{3}x$ is

If the radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $r$,then it will touch both the axes if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module