If (h, k) is the internal centre of similitud | vedclass

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

Question

(D) For the circle $x^2+y^2+2x-6y+1=0$,the centre $C_1 = (-1, 3)$ and radius $r_1 = \sqrt{(-1)^2 + 3^2 - 1} = \sqrt{1+9-1} = 3$.For the circle $x^2+y^

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) For the circle $x^2+y^2+2x-6y+1=0$,the centre $C_1 = (-1, 3)$ and radius $r_1 = \sqrt{(-1)^2 + 3^2 - 1} = \sqrt{1+9-1} = 3$.For the circle $x^2+y^2-4x+2y+4=0$,the centre $C_2 = (2, -1)$ and radius $r_2 = \sqrt{2^2 + (-1)^2 - 4} = \sqrt{4+1-4} = 1$.The internal centre of similitude $(h, k)$ divides the line segment joining the centres $C_1$ and $C_2$ internally in the ratio $r_1 : r_2 = 3 : 1$.Using the section formula,$(h, k) = \left( \frac{r_1 x_2 + r_2 x_1}{r_1 + r_2}, \frac{r_1 y_2 + r_2 y_1}{r_1 + r_2} \right)$.$(h, k) = \left( \frac{3(2) + 1(-1)}{3+1}, \frac{3(-1) + 1(3)}{3+1} \right) = \left( \frac{6-1}{4}, \frac{-3+3}{4} \right) = \left( \frac{5}{4}, 0 \right)$.Thus,$h = \frac{5}{4}$,which implies $4h = 5$.

If $(h, k)$ is the internal centre of similitude of the circles $x^2+y^2+2x-6y+1=0$ and $x^2+y^2-4x+2y+4=0$,then $4h=$

Similar Questions

The centre of the circle passing through the point $(1,1)$ and orthogonal to the circles $x^2+y^2+3x-5y+7=0$ and $x^2+y^2-6x-10y+9=0$ is

If $m_1$ and $m_2$ are the slopes of the direct common tangents drawn to the circles $x^2+y^2-2x-8y+8=0$ and $x^2+y^2-8x+15=0$,then $m_1+m_2=$

The radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$. Then

The coordinates of the centre of the circle which bisects the circumferences of the circles $x^2 + y^2 = 1$,$x^2 + y^2 + 2x - 3 = 0$,and $x^2 + y^2 + 2y - 3 = 0$ are

If the radical centre of the circles $x^2+y^2-8x-2y+8=0$,$x^2+y^2+6x+8y-24=0$,and $x^2+y^2-2x+2y+2=0$ is $(a, b)$,then $a+b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module