The two circles x^2 + y^2 - 2x + 6y + 6 = 0 a | vedclass

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

Question

(C) Given equations of the circles:$x^2 + y^2 - 2x + 6y + 6 = 0$ $(i)$$x^2 + y^2 - 5x + 6y + 15 = 0$ $(ii)$Comparing with the standard equation $x^2 +

Vedclass · Accepted Answer

Touch internally

Vedclass · Answer

(C) Given equations of the circles:$x^2 + y^2 - 2x + 6y + 6 = 0$ $(i)$$x^2 + y^2 - 5x + 6y + 15 = 0$ $(ii)$Comparing with the standard equation $x^2 + y^2 + 2gx + 2fy + c = 0$:For circle $(i)$: $g = -1, f = 3, c = 6$. Centre $A = (1, -3)$,radius $r_1 = \sqrt{(-1)^2 + 3^2 - 6} = \sqrt{1 + 9 - 6} = 2$.For circle $(ii)$: $g = -2.5, f = 3, c = 15$. Centre $B = (2.5, -3)$,radius $r_2 = \sqrt{(-2.5)^2 + 3^2 - 15} = \sqrt{6.25 + 9 - 15} = \sqrt{0.25} = 0.5$.Distance between centres $A$ and $B$ is $d = \sqrt{(2.5 - 1)^2 + (-3 - (-3))^2} = \sqrt{1.5^2 + 0^2} = 1.5$.Difference of radii is $|r_1 - r_2| = |2 - 0.5| = 1.5$.Since $d = |r_1 - r_2|$,the circles touch each other internally.

The two circles $x^2 + y^2 - 2x + 6y + 6 = 0$ and $x^2 + y^2 - 5x + 6y + 15 = 0$:

Similar Questions

Find the radius of a circle that touches the $y-$axis at point $P(0,2)$ and touches the circle $x^2 + y^2 = 16$ internally.

The point on the circle $x^2+y^2=4$ whose distance from the line $4x+3y-12=0$ is $4/5$ units is equal to

If the power of a point $(4, 2)$ with respect to the circle $x^2 + y^2 - 2\alpha x + 6y + \alpha^2 - 16 = 0$ is $9$,then the sum of the lengths of all possible intercepts made by such circles on the coordinate axes is:

The area of the circle whose centre is at $(1, 2)$ and which passes through the point $(4, 6)$ is

The number of possible common tangents that can be drawn to the circles $x^2+y^2+4x-6y-3=0$ and $x^2+y^2+4x-2y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module