# How to find the standard form of the equation of the specified circle given that it Is tangent to line x+y =2 at point (4,-2) and the center is on the x-axis?

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

#### Explanation:

Let $\left({x}_{1} , 0\right)$ be the center on the x-axis & $r$ be the radius of circle then the equation of circle

${\left(x - {x}_{1}\right)}^{2} + {\left(y - 0\right)}^{2} = {r}^{2}$

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

Now, since the above circle is tangent to the line $x + y = 2$ at $\left(4 , - 2\right)$ hence the point $\left(4 , - 2\right)$ will satisfy the equation of circle

${\left(4 - {x}_{1}\right)}^{2} + {\left(- 2\right)}^{2} = {r}^{2}$

${r}^{2} = {x}_{1}^{2} - 8 {x}_{1} + 20 \setminus \ldots . . \left(1\right)$

Now, the line $x + y = 2$ is tangent to the circle hence, substituting $y = 2 - x$ in equation of circle we get

${\left(x - {x}_{1}\right)}^{2} + {\left(2 - x\right)}^{2} = {r}^{2}$

$2 {x}^{2} - 2 \left({x}_{1} + 2\right) x + {x}_{1}^{2} + 4 - {r}^{2} = 0$

The above equation will have equal roots iff discriminant ${B}^{2} - 4 A C = 0$

${\left(- 2 \left({x}_{1} + 2\right)\right)}^{2} - 4 \left(2\right) \left({x}_{1}^{2} + 4 - {r}^{2}\right) = 0$

${x}_{1}^{2} - 4 {x}_{1} + 4 - 2 {r}^{2} = 0$

substituting the value of ${r}^{2}$ from (1) in above equation, we get

${x}_{1}^{2} - 4 {x}_{1} + 4 - 2 \left({x}_{1}^{2} - 8 {x}_{1} + 20\right) = 0$

${x}_{1}^{2} - 12 {x}_{1} + 36 = 0$

${\left({x}_{1} - 6\right)}^{2} = 0$

${x}_{1} = 6 , 6$

we get sinhle real value of $x$ i.e. $x = 6$, substituting this value in (1), we get

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

${r}^{2} = 8$

hence, the equation od circle is

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

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