Question

The largest value of $\frac{y}{x}$ such that the pair (x,y) satisfies the equation ${\left(x-3\right)}^2+{\left(y-3\right)}^2=6is$

• A. $2+2\sqrt{2}$
• B. $3+2\sqrt{2}$
• C. $2\sqrt{2}$
• D. $3\sqrt{2}$

Answer 1

The correct option is B $3+2\sqrt{2}$

${\left(x-3\right)}^2+{\left(y-3\right)}^2={\left(\sqrt{\left(6\right)}\right)}^2$

$x^2+y^2-6x-6y+9+9-6=0$

$x^2+y^2-6x-6y+12=0$

$\tan \theta =\frac{y}{x}$ $\tan \beta =\frac{3}{5}=1$ $\beta ={45}^{\circ }$

$\frac{y}{x}=\tan (\alpha +\beta )$

$OA=\sqrt{{\left(3-0\right)}^2+{\left(3-0\right)}^2}=\sqrt{18}$

$OB=\sqrt{{\left(OA\right)}^2-{\left(AB\right)}^2}=\sqrt{18-6}$

$OB=\sqrt{12}$

$\tan \alpha =\frac{\sqrt{6}}{\sqrt{12}}=\frac{1}{\sqrt{2}}$

$\frac{y}{x}=\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}$