The number of common tangents to the circles | vedclass

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

Question

(C) Given equations of the circles are:$S_1: x^2+y^2+4x-6y-12=0$$S_2: x^2+y^2-8x+10y+5=0$For $S_1$,the centre $C_1 = (-2, 3)$ and the radius $r_1 = \s

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given equations of the circles are:$S_1: x^2+y^2+4x-6y-12=0$$S_2: x^2+y^2-8x+10y+5=0$For $S_1$,the centre $C_1 = (-2, 3)$ and the radius $r_1 = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.For $S_2$,the centre $C_2 = (4, -5)$ and the radius $r_2 = \sqrt{4^2 + (-5)^2 - 5} = \sqrt{16+25-5} = \sqrt{36} = 6$.The distance between the centres $C_1$ and $C_2$ is:$C_1C_2 = \sqrt{(4 - (-2))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = \sqrt{100} = 10$.Now,compare the distance $C_1C_2$ with the sum and difference of the radii:$r_1 + r_2 = 5 + 6 = 11$$|r_1 - r_2| = |5 - 6| = 1$Since $|r_1 - r_2| Therefore,the number of common tangents to the two circles is $2$.Hence,option $C$ is correct.

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

Similar Questions

$P, Q$ and $R$ are the centres and $r_1, r_2, r_3$ are the radii respectively of three co-axial circles. Then $QRr_1^2 + RP r_2^2 + PQ r_3^2$ is equal to

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 $a+b=$

The equation of the circle which passes through the point $(1, 1)$ and intersects the given circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally is:

The radical axis of the circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ is:

If 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module