A circle meets coordinate axes at 3 points and cuts equal intercepts. If it cuts a chord of length √14 unit on x + y = 1, then square of its radius is (centre lies in first quadrant)

The correct answer is 8.

Step-by-step Explanation:

Let us determine the general equation of a circle that meets the coordinate axes at exactly 3 points and cuts equal intercepts.

Since the circle meets the coordinate axes at exactly 3 points, one of these points must be the origin (0, 0). The other two points lie on the x-axis and y-axis. Let these points be represented as (a, 0) and (0, b).

The general equation of a circle passing through the origin (0, 0), (a, 0), and (0, b) is given by:
$x^2+y^2-ax-by=0$
The center of this circle is:
$C=\left(\frac{a}{2},\frac{b}{2}\right)$

We are given two conditions:
1. The circle cuts equal intercepts on the axes. This means $\left|a\right|=\left|b\right|$.
2. The center of the circle lies in the first quadrant. This implies that both the coordinates of the center must be positive, so $a>0$ and $b>0$.
Therefore, we must have $a=b$.

Thus, the equation of the circle becomes:
$x^2+y^2-ax-ay=0$
The center of the circle is $C=\left(\frac{a}{2},\frac{a}{2}\right)$, and its radius $R$ is given by:
$R=\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2}=\sqrt{\frac{a^2}{4}+\frac{a^2}{4}}=\sqrt{\frac{a^2}{2}}=\frac{a}{\sqrt{2}}$
This gives the square of the radius as:
$R^2=\frac{a^2}{2}$

Now, let $d$ be the perpendicular distance from the center $C\left(\frac{a}{2},\frac{a}{2}\right)$ to the given line $x+y-1=0$.
$d=\frac{\left|\frac{a}{2}+\frac{a}{2}-1\right|}{\sqrt{1^2+1^2}}=\frac{\left|a-1\right|}{\sqrt{2}}$
Squaring both sides, we get:
$d^2=\frac{{\left(a-1\right)}^2}{2}$

The length of the chord cut by the circle on the line is given as $\sqrt{14}$. The relationship between the radius $R$, the distance $d$, and the chord length is:
$\text{Chord Length}=2\sqrt{R^2-d^2}$
Substituting the given chord length:
$2\sqrt{R^2-d^2}=\sqrt{14}$
Squaring both sides:
$4\left(R^2-d^2\right)=14$
$R^2-d^2=\frac{7}{2}$

Substitute the values of $R^2$ and $d^2$ in terms of $a$:
$\frac{a^2}{2}-\frac{{\left(a-1\right)}^2}{2}=\frac{7}{2}$
Multiply the entire equation by 2 to clear the denominator:
$a^2-{\left(a-1\right)}^2=7$
$a^2-\left(a^2-2a+1\right)=7$
$2a-1=7$
$2a=8$
$a=4$

Now, we can find the square of the radius:
$R^2=\frac{a^2}{2}=\frac{4^2}{2}=\frac{16}{2}=8$

Thus, the square of the radius of the circle is 8.