# How to find the standard form of the equation of the specified circle given that it Is tangent to both axis and passes through ( 2 ,-1)?

May 25, 2017

${x}^{2} + {y}^{2} - 10 x + 10 y + 25 = 0$

or ${x}^{2} + {y}^{2} - 2 x + 2 y + 1 = 0$

#### Explanation:

As the circle is tangent to both the axis, its centre is equidistant from both the axis

and as it passes through $\left(2 , - 1\right)$, it lies in fourth quadrant and coordinates of centre are of type $\left(a , - a\right)$ and its equation is

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

or ${x}^{2} + {y}^{2} - 2 a x + 2 a y + {a}^{2} = 0$

and as it passes through $\left(2 , - 1\right)$, we have

${2}^{2} + {\left(- 1\right)}^{2} - 2 a \times 2 + 2 a \times \left(- 1\right) + {a}^{2} = 0$

or ${a}^{2} - 6 a + 5 = 0$

or $\left(a - 5\right) \left(a - 1\right) = 0$

Hence $a = 5$ or $a = 1$

and equation of circle could be

${x}^{2} + {y}^{2} - 10 x + 10 y + 25 = 0$ or ${x}^{2} + {y}^{2} - 2 x + 2 y + 1 = 0$

graph{(x^2+y^2-10x+10y+25)(x^2+y^2-2x+2y+1)=0 [-1.793, 8.207, -3.94, 1.06]}