The number of direct common tangents to the c | vedclass

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

Question

(C) For the first circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.For the second circle $x^2 + y^2 - 8x - 8y + 7 = 0$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) For the first circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.For the second circle $x^2 + y^2 - 8x - 8y + 7 = 0$,the center $C_2 = (-(-8/2), -(-8/2)) = (4, 4)$ and radius $r_2 = \sqrt{4^2 + 4^2 - 7} = \sqrt{16 + 16 - 7} = \sqrt{25} = 5$.The distance between the centers $C_1 C_2 = \sqrt{(4-0)^2 + (4-0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66$.Since $|r_1 - r_2| When two circles intersect,they have $2$ direct common tangents and $0$ transverse common tangents.Therefore,the number of direct common tangents is $2$.

The number of direct common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 8x - 8y + 7 = 0$ is:

Similar Questions

If the circles $x^2+y^2+2kx-4y+1=0$ and $x^2+y^2-8x-12y+43=0$ touch each other,then $k=$

The equation of the circle passing through the origin,having its center on the line $x + y = 4$,and intersecting the circle $x^2 + y^2 - 4x + 2y + 4 = 0$ orthogonally is:

If the circle $x^2 + y^2 = 4$ bisects the circumference of the circle $x^2 + y^2 - 2x + 6y + a = 0$,then $a$ equals

If the centre of a circle,which passes through the points of intersection of the circles $x^2 + y^2 - 6x + 2y + 4 = 0$ and $x^2 + y^2 + 2x - 4y - 6 = 0$,lies on the line $y = x$,then the equation of the circle is:

The coordinates of the radical centre of the three circles $x^2 + y^2 - 4x - 2y + 6 = 0$,$x^2 + y^2 - 2x - 4y - 1 = 0$,and $x^2 + y^2 - 12x + 2y + 30 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module