The ratio of the areas of the greatest and th | vedclass

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

Question

(C) The given circles are $(x-1)^2+(y-1)^2=1$,$(x+1)^2+(y-1)^2=1$,$(x-1)^2+(y+1)^2=1$,and $(x+1)^2+(y+1)^2=1$. These circles have radius $1$ and cente

Vedclass · Accepted Answer

$\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

Vedclass · Answer

(C) The given circles are $(x-1)^2+(y-1)^2=1$,$(x+1)^2+(y-1)^2=1$,$(x-1)^2+(y+1)^2=1$,and $(x+1)^2+(y+1)^2=1$. These circles have radius $1$ and centers at $(\pm 1, \pm 1)$.The distance of the centers from the origin $(0,0)$ is $\sqrt{1^2+1^2} = \sqrt{2}$.The smallest circle is centered at the origin and touches these four circles. Its radius $r$ is the distance from the origin to the center of any of the given circles minus the radius of those circles: $r = \sqrt{2} - 1$.The greatest circle is centered at the origin and encloses these four circles. Its radius $R$ is the distance from the origin to the center of any of the given circles plus the radius of those circles: $R = \sqrt{2} + 1$.The ratio of the areas is $\frac{\pi R^2}{\pi r^2} = \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}$.

The ratio of the areas of the greatest and the smallest circles touching $(x \pm 1)^2 + (y \pm 1)^2 = 1$ is

Similar Questions

If the circles $x^{2}+y^{2}+2x+2ky+6=0$ and $x^{2}+y^{2}+2ky+k=0$ intersect orthogonally, then $k$ is equal to

Suppose the circle $S: x^2+y^2+2gx+2fy+c=0$ cuts orthogonally the two circles $S': x^2+y^2-4x-6y+11=0$ and $S'': x^2+y^2-10x-4y+21=0$. If the centre of $S=0$ lies on the bisector of the angle between the positive coordinate axes,then $2g+2f+c=$

The equation of the circle which cuts the circles $S_1 \equiv x^2+y^2-4=0$,$S_2 \equiv x^2+y^2-6x-8y+10=0$,and $S_3 \equiv x^2+y^2+2x-4y-2=0$ at the extremities of the diameters of these circles is:

If the circles $C_1: x^2+y^2+2x+4y-20=0$ and $C_2: x^2+y^2+6x-8y+9=0$ have $n$ common tangents and the length of the tangent drawn from the centre of similitude to the circle $C_2$ is $l$,then $\frac{l}{n^2} =$

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module