If all the circles of the type $x^{2} + {(y - a)}^{2} = K$
Where $a > 0.5, a \in R$ is a parameter, shares exactly two distinct points with the curve $y = x^{2}$ such that no part of the circle is outside, $y = x^{2} .$ If $K = f (a)$

List- I

List – II
A) $f (1)$ P) 1
B) $f (\frac{2}{3})$ Q) $\frac{9}{4}$
C) $f (2)$ R) $\frac{7}{4}$
D) $f (\frac{5}{4})$ S) $\frac{5}{12}$
T) $\frac{3}{4}$

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$A - P, B - Q, C - R, D - T$

$A - S, B - T, C - R, D - P$

$A - S, B - T, C - P, D - R$

$A - T, B - S, C - R, D - P$

answer is D.

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Detailed Solution

$\begin{array}{l} y + {(y - a)}^{2} = k \Rightarrow y^{2} - (2 a - 1) y + (a^{2} - k) = 0 \\ Δ = 0 \Rightarrow {(2 a - 1)}^{2} - 4 (a^{2} - k) = 0 \Rightarrow 4 k - 4 a + 1 = 0 \Rightarrow k = \frac{4 a - 1}{4} \\ f (1) = \frac{3}{4} | f (\frac{2}{3}) = \frac{\frac{8}{3} - 1}{4} = \frac{5}{12} | | f (2) = \frac{7}{4} | | f (\frac{5}{4}) = 1 | \\ \Rightarrow f (a) = \frac{4 a - 1}{4} \end{array}$