The number of common tangents of the circles | vedclass

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

Question

(C) The given circles are $C_1: x^2 + y^2 - 8x - 2y + 1 = 0$ and $C_2: x^2 + y^2 + 6x + 8y = 0$. For $C_1$,the center is $(4, 1)$ and the radius $r_1

Vedclass · Accepted Answer

two

Vedclass · Answer

(C) The given circles are $C_1: x^2 + y^2 - 8x - 2y + 1 = 0$ and $C_2: x^2 + y^2 + 6x + 8y = 0$. For $C_1$,the center is $(4, 1)$ and the radius $r_1 = \sqrt{4^2 + 1^2 - 1} = \sqrt{16} = 4$. For $C_2$,the center is $(-3, -4)$ and the radius $r_2 = \sqrt{(-3)^2 + (-4)^2 - 0} = \sqrt{25} = 5$. The distance between the centers $d = \sqrt{(4 - (-3))^2 + (1 - (-4))^2} = \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74}$. Since $\sqrt{49} Also,$r_1 + r_2 = 4 + 5 = 9$ and $|r_1 - r_2| = |4 - 5| = 1$. Since $|r_1 - r_2| Therefore,the number of common tangents is $2$.

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

Similar Questions

The line $y=mx+c$ intersects the circle $x^2+y^2=r^2$ at two distinct points if:

If $(x, 3)$ and $(3, 5)$ are the extremities of a diameter of a circle with centre at $(2, y)$,then the values of $x$ and $y$ are:

$A$ rectangle $R$ with end points of one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2x - y + 4 = 0$,then the area of $R$ is

If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangent to a circle,then the radius of the circle is

Which of the following statements are true and which are false? In each case,give a valid reason for your answer.
$q$: The centre of a circle bisects each chord of the circle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module