Question

Consider the equation of a circle x^2-2x+y^2-4y-4=0. If the line 2x-y+a=0is its diameter. Then find the value of a.

Answer 1

$\large\text{Topic: [Equation of a circle]}$

We know that a circle centered at $\large(a,b)$ with a radius of $\larger$ is -

$\text{ \cdots\longrightarrow\boxed{(x-a)^{2}+(y-b)^{2}=r^{2}.} }$

$\Large\text{\underline{\underline{Explanation}}}$

The given circle is in the form of -

$\text{ \cdots\longrightarrow\boxed{(x-1)^{2}+(y-2)^{2}=5.} }$

Hence, the circle is centered at $\large(1,2)$.

The line passes the given point.

So the linear equation has a solution pair of $\largex=1,\ y=2$.

$\cdots\longrightarrow2x-y+a=0$

$\cdots\longrightarrow2\times(1)-1\times(2)+a=0$

$\cdots\longrightarrow\boxed{a=0}$

$\large\text{[Final answer]}$

Hence, -

$\large\cdots\longrightarrow\boxed{a=0.}$