A circle {C_1} of radius 2 touches both x - a | vedclass

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

Question

(b) First circle touches both axes and radius is $2$ unit.

Hence centre of circle is $(2, 2)$.

Let radius of other circle be a and this circle a

Vedclass · Accepted Answer

$6 + 4\sqrt 2 $

Vedclass · Answer

(b) First circle touches both axes and radius is $2$ unit.

Hence centre of circle is $(2, 2)$.

Let radius of other circle be a and this circle also touches both the axes.

Hence centre of circle is $(a, a).$

This circle touches first circle

Hence, $\sqrt {{{(a - 2)}^2} + {{(a - 2)}^2}} = a + 2$ squaring both the sides,${a^2} - 12a + 4 = 0$

$a = \frac{{12 \pm \sqrt {{{(12)}^2} - 4 	imes 4 	imes 1} }}{2}$

$ = \frac{{12 \pm \sqrt {128} }}{2} = 6 \pm 4\sqrt 2 $

But $a > 2$, Hence $a = 6 - 4\sqrt 2 $ is neglected

Hence, $a = 6 + 4\sqrt 2 $.

A circle ${C_1}$ of radius $2$ touches both $x$ - axis and $y$ - axis. Another circle ${C_2}$ whose radius is greater than $2$ touches circle ${C_1}$ and both the axes. Then the radius of circle ${C_2}$ is

Similar Questions

Let the equation $x^{2}+y^{2}+p x+(1-p) y+5=0$ represent circles of varying radius $\mathrm{r} \in(0,5]$. Then the number of elements in the set $S=\left\{q: q=p^{2}\right.$ and $\mathrm{q}$ is an integer $\}$ is ..... .

The circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$ having its centre on the line, $2 x-3 y+12=0$, also passes through the point

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

If the variable line $3 x+4 y=\alpha$ lies between the two circles $(x-1)^{2}+(y-1)^{2}=1$ and $(x-9)^{2}+(y-1)^{2}=4$ without intercepting a chord on either circle, then the sum of all the integral values of $\alpha$ is .... .

If the circles ${x^2} + {y^2} - 9 = 0$ and ${x^2} + {y^2} + 2ax + 2y + 1 = 0$ touch each other, then $a =$