Circles
Circle
star_batch_jee_advanced_2025
Grade 11
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:
$(4, 0)$
$(4, 2)$
$(6, 0)$
$(7, 9)$
Step-by-Step Solution
Key Concept: The center of a circle lies on the angle bisector of any two tangent lines drawn from an external point.
The angle bisector of the angle between two tangents from an external point passes through the circle's center. Since the angle bisector of the given tangents is the $x$-axis, the center must lie on the $x$-axis at $(a,0)$. Using the distance formula from center to each tangent line: $\frac{|a-1|}{\sqrt{2}} = \sqrt{(a-3)^2 + (0-\sqrt{7}/2)^2}$. Squaring and simplifying: $(a-1)^2 = 2[(a-3)^2 + 7/2]$ yields $a = 6$ or $a = 4$.
Correct Answer: 1,3