If the two circles (x - 1)^2 + (y - 3)^2 = r^ | vedclass

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

Question

(B) For the first circle $(x - 1)^2 + (y - 3)^2 = r^2$,the center $C_1 = (1, 3)$ and radius $R_1 = r$.For the second circle $x^2 + y^2 - 8x + 2y + 8 =

Vedclass · Accepted Answer

$2 < r < 8$

Vedclass · Answer

(B) For the first circle $(x - 1)^2 + (y - 3)^2 = r^2$,the center $C_1 = (1, 3)$ and radius $R_1 = r$.For the second circle $x^2 + y^2 - 8x + 2y + 8 = 0$,the center $C_2 = (-(-8)/2, -(2)/2) = (4, -1)$ and radius $R_2 = \sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + (1)^2 - 8} = \sqrt{16 + 1 - 8} = \sqrt{9} = 3$.The distance between the centers $d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.Two circles intersect at two distinct points if $|R_1 - R_2| Substituting the values: $|r - 3| From $r + 3 > 5$,we get $r > 2$.From $|r - 3| Combining $r > 2$ and $r < 8$,we get $2 < r < 8$.

If the two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect at two distinct points,then:

Similar Questions

The equation of a circle passing through the point $(2,8)$,touching the lines $4x-3y-24=0$ and $4x+3y-42=0$,and having the $x$-coordinate of its centre less than or equal to $8$ is

For the circle $x^2+y^2-9=0$,find the equation of the chord having $(1,2)$ as its mid-point.

If one end of the diameter of the circle $x^2+y^2-4x-6y+11=0$ is $(3,4)$,then the other end of the diameter 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:

The radius of a circle whose center is $(2, 1)$ and one of its chords is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module