# A circle has its center at (2,2) and is tangent to both the x-axis and y-axis. A line tangent to this circle intersects the x-axis at (a,0) and the y-axis at (0,b). If the shaded area is equal to the area of the circle, then a + b = ? (EXACT ANSWER).

Dec 17, 2017

$a + b = \frac{a b}{4} + 2$

#### Explanation:

A circle has its center at $\left(2 , 2\right)$ and is tangent to both $x$-axis and $y$-axis, then its radius would be $2$ and equation is

${\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 4$

The circle appears as

graph{(x-2)^2+(y-2)^2=4 [-2.73, 7.27, -0.46, 4.54]}

As the liine tangent to this circle intersects the $x$-axis at $\left(a , 0\right)$ and the $y$-axis at $\left(0 , b\right)$, theequaation of tangent is

$\frac{x - 0}{y - b} = \frac{a - 0}{0 - b}$

or $\frac{x}{y - b} = - \frac{a}{b}$

or $b x = - a y + a b$

or $a y + b x - a b = 0$

its distance from center $\left(2 , , 2\right)$ is $2$

$\frac{a \times 2 + b \times 2 - a b}{\sqrt{{a}^{2} + {b}^{2}}} = 2$

and squaring

${\left(2 a + 2 b - a b\right)}^{2} = 4 \left({a}^{2} + {b}^{2}\right)$

or $4 {a}^{2} + 4 {b}^{2} + {a}^{2} {b}^{2} - 4 {a}^{2} b - 4 a {b}^{2} + 8 a b = 4 {a}^{2} + 4 {b}^{2}$

or ${a}^{2} {b}^{2} - 4 {a}^{2} b - 4 a {b}^{2} + 8 a b = 0$

as $a b \ne 0$, dividing by $a b$ we get

$a b - 4 a - 4 b + 8 = 0$

or $4 a + 4 b = a b + 8$

or $a + b = \frac{a b}{4} + 2$