If (a, b) and (c, d) are the internal and ext | vedclass

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

Question

(C) Given equations of the two circles are:$x^2+y^2+4x-5=0 \Rightarrow (x+2)^2+y^2=3^2$$x^2+y^2-6y+8=0 \Rightarrow x^2+(y-3)^2=1$Thus,$C_1=(-2, 0), r_

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Given equations of the two circles are:$x^2+y^2+4x-5=0 \Rightarrow (x+2)^2+y^2=3^2$$x^2+y^2-6y+8=0 \Rightarrow x^2+(y-3)^2=1$Thus,$C_1=(-2, 0), r_1=3$ and $C_2=(0, 3), r_2=1$.The internal centre of similitude is given by $\left(\frac{r_2x_1+r_1x_2}{r_1+r_2}, \frac{r_2y_1+r_1y_2}{r_1+r_2}ight)$:$(a, b) = \left(\frac{1(-2)+3(0)}{3+1}, \frac{1(0)+3(3)}{3+1}ight) = \left(-\frac{2}{4}, \frac{9}{4}ight) = \left(-\frac{1}{2}, \frac{9}{4}ight)$.The external centre of similitude is given by $\left(\frac{r_1x_2-r_2x_1}{r_1-r_2}, \frac{r_1y_2-r_2y_1}{r_1-r_2}ight)$:$(c, d) = \left(\frac{3(0)-1(-2)}{3-1}, \frac{3(3)-1(0)}{3-1}ight) = \left(\frac{2}{2}, \frac{9}{2}ight) = \left(1, \frac{9}{2}ight)$.Now,calculating $(a+d)(b+c)$:$(a+d)(b+c) = \left(-\frac{1}{2} + \frac{9}{2}ight) \left(\frac{9}{4} + 1ight) = \left(\frac{8}{2}ight) \left(\frac{13}{4}ight) = 4 	imes \frac{13}{4} = 13$.

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 $(a+d)(b+c)=$

Similar Questions

$A$ circle $C$ passing through the point $(1, 1)$ bisects the circumference of the circle $x^2+y^2-2x=0$. If $C$ is orthogonal to the circle $x^2+y^2+2y-3=0$,then the centre of the circle $C$ is

The circles $x^2 + y^2 + 2x - 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ touch each other:

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{3}$,then a value of $k$ is

The radical centre of the circles $x^2 + y^2 - 16x + 60 = 0$,$x^2 + y^2 - 12x + 27 = 0$,and $x^2 + y^2 - 12y + 8 = 0$ is

The number of common tangents to the circles $x^2+y^2+4x-6y-12=0$ and $x^2+y^2-8x+10y+5=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module