Question

Let a circle \(C\) have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of \(C\) on the line \(x+y=1\) is \(\sqrt{14}\), then the square of the radius of \(C\) is _____.}

Answer 1

Correct Answer: 7

Step 1: Understanding the Concept:
A circle cutting equal intercepts on both axes must have its center on the line \(y = x\) or \(y = -x\). Since the center is in the first quadrant, it lies on \(y = x\). For a circle to intersect the coordinate axes at exactly three points, it must pass through the origin \((0, 0)\) and intersect the \(x\) and \(y\) axes at one other point each.
Step 2: Key Formula or Approach:
1. Center \((h, h)\). Passes through \((0, 0) \implies r^2 = h^2 + h^2 = 2h^2\).
2. Chord length \(L = 2\sqrt{r^2 - d^2}\), where \(d\) is the distance from the center to the line.
Step 3: Detailed Explanation:
Let the center be \(O(h, h)\) and radius be \(r\). Since it passes through origin, the equation is:
\[ (x-h)^2 + (y-h)^2 = 2h^2 \implies r^2 = 2h^2 \]
Calculate the distance \(d\) from \((h, h)\) to the line \(x+y-1=0\):
\[ d = \frac{|h + h - 1|}{\sqrt{1^2 + 1^2}} = \frac{|2h - 1|}{\sqrt{2}} \]
Given chord length is \(\sqrt{14}\), so half-length is \(\frac{\sqrt{14}}{2}\).
Using Pythagoras theorem in the circle-chord triangle:
\[ \left( \frac{\sqrt{14}}{2} \right)^2 = r^2 - d^2 \]
\[ \frac{14}{4} = 2h^2 - \frac{(2h-1)^2}{2} \]
\[ 3.5 = 2h^2 - \frac{4h^2 - 4h + 1}{2} \]
\[ 7 = 4h^2 - (4h^2 - 4h + 1) \]
\[ 7 = 4h - 1 \implies 4h = 8 \implies h = 2 \].
Square of the radius \(r^2 = 2h^2 = 2(2^2) = 8\).
Step 4: Final Answer:
The square of the radius of the circle is 8.