CircleHardBloom L3
Question
If a circle passes through the point $\left( 3,\sqrt{\frac{7}{2}} \right)$ and touches $x + y = 1$ and $x - y = 1$, then the centre of the circle is at:
Options
A.$(4,0)$
B.$(4,2)$
C.$(6,0)$
D.$(7,0)$
Solution
[Key Concept: A circle touching both $x + y = 1$ and $x - y = 1$ has its centre on the bisector (the $x$-axis); equal distances to the lines and through the given point fix it.]
$$\begin{matrix} (x - 3)^{2} + \left( y - \sqrt{\frac{7}{2}} \right)^{2} = \left( \frac{x + y - 1}{\sqrt{2}} \right)^{2} \\ = \left( \frac{x - y - 1}{\sqrt{2}} \right)^{2} \end{matrix}$$
From $x + y - 1 = \pm (x - y - 1)$
$\Rightarrow \ y = 0$ or $x = 1$ (rejected)
$${(x - 3)^{2} + \frac{7}{2} = \frac{(x - 1)^{2}}{2} }{\Rightarrow x = 4,6 }$$
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