Question

Find the equation of a circle which touches the lines $7x^2-18xy+7y^2=0$ and the circle $x^2+y^2-8x-8y=0$ and is contained in the given circle.

Answer 1

Lines $\Rightarrow 7x^2-18xy+7y^2$ ........(1)

Coefficient of $x=y\Rightarrow$ both lines through origin

- center of circle is on angle bisector of lines.

- Angle bisectors are ether $(x+y)$ or $(x-y)$

$\Rightarrow$ If is (x-y), as the$\underset{\underset{x^2+y^2-8x-8y=u}{\downarrow }}{contained}$ circle lies in $1st$ quadrant

$\Rightarrow$ so center lies on $y=x,\Rightarrow (a,a)$

$(x-a{)}^2+(y-a{)}^2=c^2$ ..........(2)

$1st$ distance from center to tangent = Radius

Let tangent $\Rightarrow y=mx\Rightarrow \frac{|y-m^2|}{\sqrt{1+m^2}}=R\Rightarrow \frac{a-ma}{\sqrt{1+m^2}}=R$

$\Rightarrow R^2(m^2+1)=a^2(1+m^2-2m)\Rightarrow R^2{m}^2+R^2=a^2+a^2{m}^2-2m{a}^2$

$\Rightarrow m^2(R^2-a^2)=a^2-R^2-2m{a}^2$

$m^2=\frac{(a+R)(a+R)}{(R+a)(R-a)}-\frac{2m{a}^2}{(R^2-a^2)}$

Company with (1) sum of roots.

$\frac{2a^2}{R^2-a^2}=\frac{-18}{7}\Rightarrow 14a^2=-18R^2+18a^2\Rightarrow 4a^2=18R^2$

$\Rightarrow a=\frac{3}{\sqrt{2}}Ror\frac{3}{\sqrt{2}}C$

Radius of given $\Rightarrow (x-y{)}^2(y-y{)}^2=(4{\sqrt{)}}^2$

$R-R_1=C{C}_1$

$\Rightarrow (4\sqrt{2}-R{)}^2={(4-\frac{3R}{\sqrt{2}})}^2\times 2\Rightarrow 4\sqrt{2}-R=-\sqrt{2}\left(4\frac{-3R}{\sqrt{2}}\right)$

$\Rightarrow 8\sqrt{2}=4R\Rightarrow R=2\sqrt{2}\Rightarrow a=\frac{3}{\sqrt{4}}\times 2\sqrt{2}\Rightarrow a=6$

Center =(6,6). Radius =2\sqrt 2

Circle equation

$\Rightarrow (x-6{)}^2+(y-6{)}^2=(2\sqrt{2}{)}^2$