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

(D) For the first circle ${x^2} + {y^2} - 4x - 6y - 12 = 0$,the center $c_{1} = (2, 3)$ and radius $r_{1} = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 1

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) For the first circle ${x^2} + {y^2} - 4x - 6y - 12 = 0$,the center $c_{1} = (2, 3)$ and radius $r_{1} = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.For the second circle ${x^2} + {y^2} + 6x + 18y + 26 = 0$,the center $c_{2} = (-3, -9)$ and radius $r_{2} = \sqrt{(-3)^2 + (-9)^2 - 26} = \sqrt{9 + 81 - 26} = \sqrt{64} = 8$.The distance between the centers $c_{1}c_{2} = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.Since $c_{1}c_{2} = r_{1} + r_{2} = 5 + 8 = 13$,the two circles touch each other externally.When two circles touch each other externally,the number of common tangents is $3$.

List-$I$	List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$	$I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$	$II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$	$III.$ $60^{\circ}$
	$IV.$ $30^{\circ}$

The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is

Similar Questions

The equation of the circle having the chord $x \cos \alpha + y \sin \alpha = p$ of the circle $x^2 + y^2 = a^2$ as its diameter is:

Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 - 2x - 2y - 2 = 0$ and $x^2 + y^2 - 6x - 6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles,then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is ............. $sq. \, units$.

The circles $x^2 + y^2 + 4x + d = 0$ and $x^2 + y^2 + 4fy + d = 0$ touch each other if:

The equation of the circle which cuts the circles $S_1 \equiv x^2+y^2-4=0$,$S_2 \equiv x^2+y^2-6x-8y+10=0$,and $S_3 \equiv x^2+y^2+2x-4y-2=0$ at the extremities of the diameters of these circles is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module