If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then is equal to [2025]

Question

If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then $10 k + r^{2}$ is equal to [2025]

Smart Achievers · Accepted Answer

(4)

Let $P \equiv (4, 6), Q \equiv (- 1, 5), R \equiv (0, 0)$ and $S \equiv (k, 3 k)$ .

Slope of line through point P and Q is

$m_{1} = \frac{1}{5}$ ... (i)

Slope of line through point Q and R is

$m_{2} = - 5$ ... (ii)

From (i) and (ii), we get

$m_{1} m_{2} = - 1$

So, line passes by P and Q is perpendicular to line passes by Q and R.

So, P and R will represent end point of diameter of circle so we have

(x – 4)(x – 0) + (y – 6) (y – 0) = 0

$\Rightarrow x^{2} + y^{2} - 4 x - 6 y = 0$

Since (k, 3k) lies on it

$∴ k^{2} + 9 k^{2} - 4 k - 18 k = 0$

$\Rightarrow 10 k^{2} - 22 k = 0$

$\Rightarrow$ k = 0, $\frac{11}{5}$

Since, k = 0 is not possible, so $k = \frac{11}{5}$

Also, $r = \frac{\sqrt{4^{2} + 6^{2}}}{2} = \frac{\sqrt{52}}{2} = \sqrt{13}$

$∴ 10 k + r^{2} = 10 \times \frac{11}{5} + {(\sqrt{13})}^{2} = 35$ .

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