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

(C) For the circle $x^2+y^2+2x+8y-23=0$,the center $C_1 = (-1, -4)$ and radius $r_1 = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23} = \sqrt{40}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For the circle $x^2+y^2+2x+8y-23=0$,the center $C_1 = (-1, -4)$ and radius $r_1 = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23} = \sqrt{40} = 2\sqrt{10}$.For the circle $x^2+y^2-4x-10y+19=0$,the center $C_2 = (2, 5)$ and radius $r_2 = \sqrt{(2)^2 + (5)^2 - 19} = \sqrt{4 + 25 - 19} = \sqrt{10}$.The distance between the centers $d = C_1C_2 = \sqrt{(2 - (-1))^2 + (5 - (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$.Since $r_1 + r_2 = 2\sqrt{10} + \sqrt{10} = 3\sqrt{10}$,we have $d = r_1 + r_2$.Because the distance between the centers is equal to the sum of the radii,the two circles touch each other externally.When two circles touch each other externally,they have exactly $3$ common tangents.

The number of common tangents to the circles $x^2+y^2+2x+8y-23=0$ and $x^2+y^2-4x-10y+19=0$ is:

Similar Questions

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

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 . . . . . .

If the product of the lengths of the perpendiculars drawn from the ends of a diameter of the circle $x^2+y^2=4$ onto the line $x+y+1=0$ is maximum,then the two ends of that diameter are

Suppose that the earth is a sphere of radius $6400 \, km$. The height from the earth's surface from where exactly a fourth of the earth's surface is visible,is $...... \, km$.

$A$ line meets the circle $x^2+y^2-4x-4y-8=0$ at two points $A$ and $B$. If $P(2,-2)$ is a point on the circle such that $PA=PB=2$,then the equation of the line $AB$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module