If T_1 T_1^{prime} and T_2 T_2^{prime} are th | vedclass

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

Question

(D) For the circle $S = x^2 + y^2 - 2x - 4y - 4 = 0$, the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2 + 2^2 - (-4)} = \sqrt{9} = 3$.For the circ

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) For the circle $S = x^2 + y^2 - 2x - 4y - 4 = 0$, the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2 + 2^2 - (-4)} = \sqrt{9} = 3$.For the circle $S^{\prime} = x^2 + y^2 + 4x + 4y + 4 = 0$, the center $C_2 = (-2, -2)$ and radius $r_2 = \sqrt{(-2)^2 + (-2)^2 - 4} = \sqrt{4} = 2$.The distance between the centers $C_1 C_2 = \sqrt{(1 - (-2))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.The length of the direct common tangent is given by $L = \sqrt{(C_1 C_2)^2 - (r_1 - r_2)^2}$.$L = \sqrt{5^2 - (3 - 2)^2} = \sqrt{25 - 1} = \sqrt{24} = 2 \sqrt{6}$.Since $T_1 T_1^{\prime}$ represents the length of the common tangent, the distance is $2 \sqrt{6}$.

Similar Questions

If $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p$,where $0 < p < a$,then $\lambda=$

Let $x+y=0$ be the radical axis of the circles $S \equiv x^2+y^2+2gx+2fy+c=0$ and $S' \equiv x^2+y^2-6x-4y+4=0$. If the radius of the circle $S=0$ is $1$,then find the value of $g+f$.

The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is

If the circles $x^2+y^2-4x+6y+13-a^2=0$ and $x^2+y^2-10x-2y+17=0$ intersect in two distinct points,then '$a$' is

If the circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts the three circles $x^{2}+y^{2}-5=0$, $x^{2}+y^{2}-8x-6y+10=0$, and $x^{2}+y^{2}-4x+2y-2=0$ at the extremities of their diameters, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module