Let the set of all values of r, for which the | vedclass

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

Question

(A) The first circle is $C_1: (x+1)^2+(y+4)^2=r^2$ with center $O_1(-1, -4)$ and radius $r_1 = |r|$.The second circle is $C_2: x^2+y^2-4x-2y-4=0$. Rew

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The first circle is $C_1: (x+1)^2+(y+4)^2=r^2$ with center $O_1(-1, -4)$ and radius $r_1 = |r|$.The second circle is $C_2: x^2+y^2-4x-2y-4=0$. Rewriting in standard form: $(x-2)^2+(y-1)^2 = 4+1+4 = 9$, so center $O_2(2, 1)$ and radius $r_2 = 3$.For two circles to intersect at two distinct points, the distance between their centers $d = O_1O_2$ must satisfy $|r_1 - r_2| Calculate $d = \sqrt{(2 - (-1))^2 + (1 - (-4))^2} = \sqrt{3^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$.Thus, $|r - 3| From $|r - 3| From $\sqrt{34}  \sqrt{34} - 3$. Since $\sqrt{34} \approx 5.83$, $\sqrt{34} - 3 > 0$, so $r > \sqrt{34} - 3$ or $r Since $r$ is a radius, $r > 0$. The intersection of $r \in (3 - \sqrt{34}, 3 + \sqrt{34})$ and $r > \sqrt{34} - 3$ is $r \in (\sqrt{34} - 3, \sqrt{34} + 3)$.Thus, $\alpha = \sqrt{34} - 3$ and $\beta = \sqrt{34} + 3$.$\alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) = 34 - 9 = 25$.

Let the set of all values of $r$, for which the circles $(x+1)^{2}+(y+4)^{2}=r^{2}$ and $x^{2}+y^{2}-4x-2y-4=0$ intersect at two distinct points, be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

Similar Questions

If the radical centre of the circles $x^2+y^2-8x-2y+8=0$,$x^2+y^2+6x+8y-24=0$,and $x^2+y^2-2x+2y+2=0$ is $(a, b)$,then $a+b=$

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is:

If $(-1, -1)$ is the radical centre of the circles $x^2 + y^2 + 2gx - 4y + 4 = 0$,$x^2 + y^2 + 6x + 2fy + 12 = 0$,and $x^2 + y^2 + 10y + 20 = 0$,then $g - f = $

If the acute angle between the circles $S \equiv x^2+y^2+2kx+4y-3=0$ and $S' \equiv x^2+y^2-4x+2ky+9=0$ is $\cos^{-1}(\frac{3}{8})$ and the centre of $S'=0$ lies in the first quadrant,then the radical axis of $S=0$ and $S'=0$ 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