A circle meets coordinate axes at 3 points and cuts equal intercepts. If it cuts a chord of length $\sqrt{14}$ unit on $x + y = 1$, then square of its radius is (centre lies in first quadrant):
Step 1: Understanding the Concept:
A circle cutting equal intercepts on axes and passing through 3 points must pass through the origin \( (0,0) \).
Its centre must lie on the line \( y = x \) to maintain equal intercepts. Step 2: Key Formula or Approach:
The circle passes through \( (0,0), (k,0), (0,k) \).
Centre is \( C(k/2, k/2) \). Let radius be \( r \), then \( r^2 = (k/2)^2 + (k/2)^2 = k^2/2 \).
So \( k = \sqrt{2}r \). Centre is \( (\frac{r}{\sqrt{2}}, \frac{r}{\sqrt{2}}) \). Step 3: Detailed Explanation:
The perpendicular distance \( p \) from centre to the line \( x + y - 1 = 0 \) is:
\[ p = \frac{|\frac{r}{\sqrt{2}} + \frac{r}{\sqrt{2}} - 1|}{\sqrt{1^2 + 1^2}} = \frac{|\sqrt{2}r - 1|}{\sqrt{2}} \]
Chord length \( L = 2\sqrt{r^2 - p^2} \). Given \( L = \sqrt{14} \).
\[ 14 = 4 \left( r^2 - \frac{(\sqrt{2}r - 1)^2}{2} \right) \]
\[ 7 = 2 \left( r^2 - \frac{2r^2 + 1 - 2\sqrt{2}r}{2} \right) \]
\[ 7 = 2r^2 - 2r^2 - 1 + 2\sqrt{2}r \]
\[ 7 = 2\sqrt{2}r - 1 \implies 8 = 2\sqrt{2}r \implies 4 = \sqrt{2}r \]
Square both sides: \( 16 = 2r^2 \implies r^2 = 8 \). Step 4: Final Answer:
The square of the radius of the circle is 8.
