Question

The intercept on the line $y=x$ by the circle $x^2+y^2-2x=0$ is $AB$. Equation of the circle on $AB$ as a diameter is

• A. $x^2+y^2=1$
• B. $x\left(x-1\right)+y\left(y-1\right)=0$
• C. $x^2+y^2=2$
• D. $\left(x-1\right)\left(x-2\right)+\left(y-1\right)\left(y-2\right)=0$

Answer 1

The correct option is B

$x\left(x-1\right)+y\left(y-1\right)=0$

Step 1: Find intersecting points of a diameter

Given the equation of the straight line: $y=x$.

Given the equation of the circle: $x^2+y^2-2x=0$.

For the intercepts, substitute $y$ with $x$ in the equation of the circle.

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

Thus, the solutions to the quadratic equation are $x_1=0$ and $x_2=1$.

From the equation of the straight line, $y=x$ it can be written that $y_1=0$ and $y_2=1$.

Hence, the points of intersects are $A\left(0,0\right)$ and $B\left(1,1\right)$.

Step 2: Find equation of a straight line

It is known that the equation of a circle whose extremities are $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ can be given by, $\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$.

Hence, the equation of the circle can be given by,

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

Hence, the equation of the circle is $x\left(x-1\right)+y\left(y-1\right)=0$, so the correct answer is option (B).