The internal centre of similitude of the two | vedclass

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

Question

(C) The given circles are:$x^2+y^2-4x-6y+12=0 \dots(1)$Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3

Vedclass · Accepted Answer

$\left(1, \frac{5}{2}\right)$

Vedclass · Answer

(C) The given circles are:$x^2+y^2-4x-6y+12=0 \dots(1)$Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-12} = \sqrt{13-12} = 1$.$x^2+y^2+4x-2y-4=0 \dots(2)$Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_2 = (-2, 1)$ and radius $r_2 = \sqrt{(-2)^2+1^2-(-4)} = \sqrt{4+1+4} = 3$.The internal centre of similitude divides the line segment joining the centers $C_1$ and $C_2$ internally in the ratio $r_1 : r_2 = 1 : 3$.Using the section formula,the coordinates are:$C = \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}\right)$$C = \left(\frac{1(-2) + 3(2)}{1+3}, \frac{1(1) + 3(3)}{1+3}\right)$$C = \left(\frac{-2+6}{4}, \frac{1+9}{4}\right) = \left(\frac{4}{4}, \frac{10}{4}\right) = \left(1, \frac{5}{2}\right)$.

The internal centre of similitude of the two circles $x^2+y^2-4x-6y+12=0$ and $x^2+y^2+4x-2y-4=0$ is

Similar Questions

$(a, 0)$ 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:

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 $g+f$.

If $(a, b)$ and $(c, d)$ are the internal and external centres of similitude of the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-6y+8=0$ respectively,then $(a+d)(b+c)=$

If $x^2+y^2-6x-8y+12=0$ and $x^2+y^2-4x+6y+k=0$ cut orthogonally,then $k=$

If one of the diameters of the circle,given by the equation $x^2+y^2-4x+6y-12=0$,is a chord of a circle,$S$,whose centre is at $(-3,2)$,then the length of the radius of $S$ is . . . . . . units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module