Mathematics - Circle Question with Solution | TestHub

The centre of a circle passing through points M and N lies on the perpendicular bisector $\mathrm{y}=3-x$ of $MN$. Denote the centre by $C(a,3-a)$, the equation of the circle is $(x-a{)}^2+(y-3+a{)}^2=2\left(1+a^2\right)$

Since for a chord with a fixed length the angle at the circumference subtended by the corresponding arc will become larger as the radius of the circle becomes smaller. When $\angle \mathrm{MPN}$ reaches its maximum value the circle through the three points $\mathrm{M},\mathrm{N}$ and P will be tangent to the x -axis at P , which means $2\left(1+{\mathrm{a}}^2\right)=(\mathrm{a}-3{)}^2\Rightarrow \mathrm{a}=1$ or $\mathrm{a}=-7$

Thus the point of contact are $\mathrm{P}(1,0)$ or ${\mathrm{P}}^{\prime }(-7,0)$ respectively.

But the radius of circle through the points $\mathrm{M},\mathrm{N}$ and ${\mathrm{P}}^{\prime }$ is larger than that of circle through points $M,N$ and $P$.

Therefore, $\angle \mathrm{MPN}>\angle {\mathrm{MP}}^{\prime }\mathrm{N}$. Thus $\mathrm{P}=(1,0)$

$\therefore \mathrm{x}$-coordinate of $\mathrm{P}=1$.