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

(A) For the first circle $x^2 + y^2 - 8x - 2y + 1 = 0$,the center $C_1 = (4, 1)$ and radius $r_1 = \sqrt{4^2 + 1^2 - 1} = \sqrt{16 + 1 - 1} = \sqrt{16

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) For the first circle $x^2 + y^2 - 8x - 2y + 1 = 0$,the center $C_1 = (4, 1)$ and radius $r_1 = \sqrt{4^2 + 1^2 - 1} = \sqrt{16 + 1 - 1} = \sqrt{16} = 4$.For the second circle $x^2 + y^2 + 6x + y = 0$,the center $C_2 = (-3, -0.5)$ and radius $r_2 = \sqrt{(-3)^2 + (-0.5)^2 - 0} = \sqrt{9 + 0.25} = \sqrt{9.25} \approx 3.04$.The distance between the centers $d = \sqrt{(4 - (-3))^2 + (1 - (-0.5))^2} = \sqrt{7^2 + 1.5^2} = \sqrt{49 + 2.25} = \sqrt{51.25} \approx 7.16$.Since $d = \sqrt{51.25} > r_1 + r_2 = 4 + 3.04 = 7.04$,the circles lie outside each other.Therefore,the number of common tangents is $4$.

The number of common tangents to the circles $x^2 + y^2 - 8x - 2y + 1 = 0$ and $x^2 + y^2 + 6x + y = 0$ is:

Similar Questions

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k$ is equal to :

The equation of the direct common tangent of the circles $x^2+y^2-6x-4y-23=0$ and $x^2+y^2+2x+2y+1=0$ is

If the straight line $x \cos \alpha + y \sin \alpha = P$ intersects the circle $x^2 + y^2 = a^2$ at $A$ and $B$,then the equation of the circle with diameter $\overline{AB}$ is

If $x^2+y^2-4x-2y+5=0$ and $x^2+y^2-6x-4y-3=0$ are members of a coaxial system of circles,then the centre of a point circle in the system is

If $(\alpha, \beta)$ is the external centre of similitude of the circles $x^2+y^2=3$ and $x^2+y^2-2x+4y+4=0$,then $\frac{\beta}{\alpha}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module