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 lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ is:

The equation of the circle which passes through the point $(3,2)$,bisects the circumference of the circle $x^2+y^2=15$,and cuts the circle $x^2+y^2+4x+6y+3=0$ orthogonally is

In the co-axial system of circles $x^2 + y^2 + 2gx + c = 0$,where $g$ is a parameter,if $c > 0$,then the circles are

$A$ circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally,then the equation of the circle is

Two given circles $x^2 + y^2 + ax + by + c = 0$ and $x^2 + y^2 + dx + ey + f = 0$ will intersect each other orthogonally,only when

Vedclass Test Series

Exam Paper Generator

Online Exam Module