<p>If a circle passes through the point $\left(3, \frac{7}{2}\right)$ and touches $x + y = 1$ and $x - y = 1$, then the centre of the circle is at:</p>
Step-by-Step Solution
Key Concept: The centre of a circle tangent to two lines lies on the angle bisector of those lines, and must be equidistant from both tangent lines.
<p><strong>Analysis:</strong> Since the circle touches both lines $x + y = 1$ and $x - y = 1$, the centre must be equidistant from both lines. The distance from point $(h,k)$ to line $x + y - 1 = 0$ is $\frac{|h+k-1|}{\sqrt{2}}$ and to line $x - y - 1 = 0$ is $\frac{|h-k-1|}{\sqrt{2}}$. Setting these equal and considering the first quadrant constraint, we get $k = 0$. The circle passes through $\left(3, \frac{7}{2}\right)$, so $(3-h)^2 + \left(\frac{7}{2}\right)^2 = r^2$ where $r$ is the radius (distance from centre to the lines). Testing the options shows $(4,0)$ satisfies all conditions.</p><p>∴ Answer is (a).</p>
Correct Answer: a