The number of common tangents that can be dra | vedclass

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

Question

(A) For the circle $x^2+y^2=1$:Centre $C_1 = (0,0)$,Radius $R_1 = 1$.For the circle $x^2+y^2-2x-6y+6=0$:Centre $C_2 = (1,3)$,Radius $R_2 = \sqrt{1^2+3

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) For the circle $x^2+y^2=1$:Centre $C_1 = (0,0)$,Radius $R_1 = 1$.For the circle $x^2+y^2-2x-6y+6=0$:Centre $C_2 = (1,3)$,Radius $R_2 = \sqrt{1^2+3^2-6} = \sqrt{4} = 2$.The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1+9} = \sqrt{10}$.Since $\sqrt{10} \approx 3.16$ and $R_1+R_2 = 1+2 = 3$,we have $d > R_1+R_2$.Because the distance between the centres is greater than the sum of the radii,the two circles do not intersect and lie outside each other.Therefore,there are $4$ common tangents (two direct and two transverse).

The number of common tangents that can be drawn to the circles $x^2+y^2=1$ and $x^2+y^2-2x-6y+6=0$ is

Similar Questions

The radius of the circle passing through the points $(-1, 1)$,$(2, -1)$,and $(1, 0)$ is

If the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$ is $45^{\circ}$,then a value of $k$ is

The equation of the circle which passes through $(1, 0)$ and $(0, 1)$ and has its radius as small as possible,is

If the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 - 2gx + g^2 - b^2 = 0$ touch each other externally,then:

$A$ circular wire of radius $7\,cm$ is cut and bent again into an arc of a circle of radius $12\,cm$. The angle subtended by the arc at the centre is ......$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module