Question

The equation of circle which touches $X$ and $Y$ - axes at the points $(1,0)\&(0,1)$ respectively is

• A. $x^2+y^2–4y+3=0$
• B. $x^2+y^2–2y–2=0$
• C. $x^2+y^2–2x–2y+2=0$
• D. $x^2+y^2–2x–2y+1=0$

Answer 1

The correct option is D

$x^2+y^2–2x–2y+1=0$

Explanation for the correct option:

Finding the equation of circle:

The given points at which the circle touches $X$ - axis and $Y$- axis respectively

$(1,0)\&(0,1)$

Thus drawing figure according to the given information.

Therefore, from the figure, center of the circle be $(1,1)$

We know the equation of the circle with center $(a,b)$ and radius $r$

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

Thus, the circle with center $(1,1)$

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

Since it passes through $(0,1)$. Therefore,

$$\begin{gathered} \Rightarrow {(0-1)}^2+{\left(1-1\right)}^2=r^2 \\ \Rightarrow r=2 \end{gathered}$$

Substituting the value of $r$ in equation $\left(i\right)$ we have

$$\begin{gathered} \Rightarrow {(x-1)}^2+{\left(y-1\right)}^2=1^2 \\ \Rightarrow \left(x^2-2x+1\right)+\left(y^2-2y+1\right)=1 \\ \Rightarrow x^2+y^2-2x-2y+1=0 \end{gathered}$$

Hence, option (D) is the correct answer.