Let a circle C with radius r passes through four distinct points (0, 0), (K, 3k), (2, 3) and (-1, 5), such that k ≠ 0, then (10k + 2r^2) is equal to

Question

Let a circle C with radius r passes through four distinct points (0, 0), (K, 3k), (2, 3) and (-1, 5), such that \(k \neq 0,\) then \(\left(10 k+2 r^{2}\right)\) is equal to

(1) 35

(2) 34

(3) 27

(4) 32

Answer 1

Correct option is: (3) 27  

Centre \(\equiv\left(\frac{-1}{2}, \frac{5}{2}\right)\)

\(\Rightarrow\) Radius \(=\frac{\sqrt{26}}{2}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{5}{2}\right)^{2}=\frac{26}{4}\)

\((2 x+1)^{2}+(2 y-5)^{2}=26\)

\(\Rightarrow 4 x^{2}+4 y^{2}+4 x-20 y=0\)

(K, 3k) lie on circle

\(4 k^{2}+36 k^{2}+4 k-60 k=0\)

\(40 k-56=0 \Rightarrow k=\frac{7}{5}\)

\(\Rightarrow 10 k+2 r^{2}=14+13=27\)