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 $y = 2x$ is a chord of the circle $x^{2} + y^{2} = 10x$,then find the equation of the circle having this chord as its diameter.

If the radical axis of the circles $x^2+y^2+2 \alpha x+2 \beta y+c=0$ and $x^2+y^2+\frac{3}{2} x+4 y+c=0$ touches the circle $x^2+y^2+2 x+2 y+1=0$,then $4 \alpha \beta-8 \alpha-3 \beta+10=$

If the centre $(\alpha, \beta)$ of a circle cutting the circles $x^2+y^2-2y-3=0$ and $x^2+y^2+4x+3=0$ orthogonally lies on the line $2x-3y+4=0$,then $2\alpha+\beta=$

The intercept on the line $y = x$ by the circle $x^2 + y^2 - 2x = 0$ is $AB$. The equation of the circle with $AB$ as a diameter is:

The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module