# How do you write an equation for a circle tangent to the line x - y = 2 at the point (4,2) and the center is on the x-axis?

Jul 4, 2016

$= {x}^{2} + {y}^{2} - 12 x + 28 = 0$

#### Explanation:

Let the coordinate of the center of the circle lying on x-axis be $\left(a , 0\right)$ and its radius is r. The equation of the circle will be

${\left(x - a\right)}^{2} + {\left(y - 0\right)}^{2} = {r}^{2}$

${\left(x - a\right)}^{2} + {y}^{2} = {r}^{2.} \ldots . \left(1\right)$

Now the point (4,2) is lying on the circle.So

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

${a}^{2} - 8 a + 20 = {r}^{2.} \ldots . \left(2\right)$

Now it is given that $x - y = 2$ is the equation of tangent to the circle at the point(4,2) on the circle.
Witing the equation of the tangent in $y = m x + c$ form we have the equation of the tangent as $y = x - 2$,So it is obvious that the slope of the tangent is 1.
Hence the slope of the normal passing through (4.2) is $- 1$

So equation of the normal at (4,2) will be

$\left(y - 2\right) = \left(- 1\right) \left(x - 4\right)$

$\implies y - 2 = - x + 4$

$\implies x + y = 6. \ldots \left(3\right)$

Now as the center (a,0) is lying on the normal ,it will satisfy the equation of normal.
So inserting$x = a \mathmr{and} y = 0$ in (3) we get
$a + 0 = 6 \implies a = 6$

Putting this value of a =6 in (2) we get

${a}^{2} - 8 a + 20 = {r}^{2}$

${6}^{2} - 8 \times 6 + 20 = {r}^{2}$

$\implies {r}^{2} = 8$

Now finally plugging in the value of $a = 6 \mathmr{and} {r}^{2} = 8$ in equation(1) we get the required equation of circle

${\left(x - 6\right)}^{2} + {y}^{2} = 8$

$\implies {x}^{2} - 12 x + 36 + {y}^{2} = 8$

$= {x}^{2} + {y}^{2} - 12 x + 28 = 0$