If the circles (x+1)^2+(y+2)^2=r^2 and x^2+y^ | vedclass

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

Question

(C) For two circles to intersect at two distinct points,the distance between their centers $C_1C_2$ must satisfy the condition: $|r_1 - r_2| < C_1C_2

Vedclass · Accepted Answer

$3 < r < 7$

Vedclass · Answer

(C) For two circles to intersect at two distinct points,the distance between their centers $C_1C_2$ must satisfy the condition: $|r_1 - r_2| For the first circle $(x+1)^2 + (y+2)^2 = r^2$,the center $C_1 = (-1, -2)$ and radius $r_1 = r$.For the second circle $x^2 + y^2 - 4x - 4y + 4 = 0$,the center $C_2 = (2, 2)$ and radius $r_2 = \sqrt{2^2 + 2^2 - 4} = \sqrt{4+4-4} = 2$.The distance between the centers $C_1C_2 = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.Applying the condition $|r - 2| $1$) $|r - 2| $2$) $5  3$.Combining these inequalities,we get $3 < r < 7$.

If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4x-4y+4=0$ intersect at exactly two distinct points,then

Similar Questions

The equation of the circle passing through the points of intersection of the circles $x^2+y^2-2px=0$ and $x^2+y^2-2qy=0$ and having its centre on the line $\frac{x}{p}-\frac{y}{q}=2$ is:

If $P$ and $Q$ are the points of intersection of the circles $x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $x^2 + y^2 + 2x + 2y - p^2 = 0$,then for what value of $p$ does a circle pass through $P, Q$ and $(1, 1)$?

If the circles $x^2+y^2-8x-8y+28=0$ and $x^2+y^2-8x-6y+25-\alpha^2=0$ have only one common tangent,then $\alpha=$

Find the equation of the circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$.

$A$ circle $C$ passes through $(2a, 0)$ and the line $2x = a$ is the radical axis of the circle $C$ and the circle $x^2 + y^2 = a^2$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module