The distance of the origin from the external | vedclass

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

Question

(A) Given circles are $S_1: x^2+y^2-8x-10y-8=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,center $C_1 = (4, 5)$ and radius $r_1 = \sqrt{4^2+5^2-(-8)} =

Vedclass · Accepted Answer

$\frac{3 \sqrt{26}}{5}$

Vedclass · Answer

(A) Given circles are $S_1: x^2+y^2-8x-10y-8=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,center $C_1 = (4, 5)$ and radius $r_1 = \sqrt{4^2+5^2-(-8)} = \sqrt{16+25+8} = \sqrt{49} = 7$. For $S_2$,center $C_2 = (-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-(-2)} = \sqrt{1+1+2} = \sqrt{4} = 2$. The external centre of similitude $Q$ divides the line segment joining $C_1$ and $C_2$ externally in the ratio $r_1 : r_2$. $Q = \left( \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} ight) = \left( \frac{7(-1) - 2(4)}{7-2}, \frac{7(1) - 2(5)}{7-2} ight) = \left( \frac{-7-8}{5}, \frac{7-10}{5} ight) = \left( -3, -\frac{3}{5} ight)$. The distance of $Q$ from the origin $(0,0)$ is $D = \sqrt{(-3)^2 + (-3/5)^2} = \sqrt{9 + 9/25} = \sqrt{\frac{225+9}{25}} = \sqrt{\frac{234}{25}} = \frac{\sqrt{9 	imes 26}}{5} = \frac{3\sqrt{26}}{5}$.

The distance of the origin from the external centre of similitude for the circles $x^2+y^2-8x-10y-8=0$ and $x^2+y^2+2x-2y-2=0$ is

Similar Questions

If the angle between the circles $x^2+y^2-2x+2y+1=0$ and $x^2+y^2+2x-2y+k=0$ is $\frac{\pi}{3}$,then

If the circle $x^2+y^2+2kx+4y-4=0$ has its centre in the $4^{\text{th}}$ quadrant and touches the circle $x^2+y^2+6x-2y+6=0$,then $k=$

If the circles $x^2+y^2+2 \lambda x+2=0$ and $x^2+y^2+4y+2=0$ touch each other,then $\lambda=$

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is:

$A$ circle $S$ passes through the points of intersection of the circles $x^2+y^2-2x+2y-2=0$ and $x^2+y^2+2x-2y+1=0$. If the centre of this circle $S$ lies on the line $x-y+6=0$,then the radius of the circle $S$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module