Find the number of common tangents that can b | vedclass

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

Question

(C) For the first circle $x^2 + y^2 - 4x - 6y - 3 = 0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2 + 3^2 - (-3)} = \sqrt{4 + 9 + 3} = 4$.For

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For the first circle $x^2 + y^2 - 4x - 6y - 3 = 0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2 + 3^2 - (-3)} = \sqrt{4 + 9 + 3} = 4$.For the second circle $x^2 + y^2 + 2x + 2y + 1 = 0$,the center $C_2 = (-1, -1)$ and radius $r_2 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1 + 1 - 1} = 1$.The distance between the centers $C_1C_2 = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.Since $C_1C_2 = r_1 + r_2$ $(5 = 4 + 1)$,the two circles touch each other externally.Therefore,the number of common tangents is $3$.

Find the number of common tangents that can be drawn to the circles $x^2 + y^2 - 4x - 6y - 3 = 0$ and $x^2 + y^2 + 2x + 2y + 1 = 0$.

Similar Questions

$P(3,1)$,$Q(6,5)$ and $R(x,y)$ are three points such that the angle $\angle PRQ$ is a right angle and the area of the $\Delta RPQ = 5$. Then the number of such points $R$ is

The line $y=mx+c$ intercepts the circle $x^2+y^2=r^2$ in two distinct points,if

If the line $x + 2by + 7 = 0$ is a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$,then $b = $

Among all the parallelograms whose diagonals are $10$ and $4$,the one having maximum area has its perimeter lying in the interval

$A$ circle passing through the point $(1,0)$ makes an intercept of length $4$ units on the $X$-axis and an intercept of length $2\sqrt{11}$ units on the $Y$-axis. If the centre of the circle lies in the fourth quadrant,then the radius of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module