Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

EXPLANATION.

$\sf \to \: circle \: passing \: through \: point \: (0 \:, 1) \\ \\ \sf \to \: touch \: the \: curve \: \implies \: y \: = {x}^{2} \: \: \: at \: \: \: (2,4) \: \: \: is$

$\sf \to \: let \: we \: assume \: that \: centre \: of \: circle \: = (h,k) \\ \\ \sf \to \: oa {}^{2} = {ob}^{2} \\ \\ \sf \to \: as \: we \: know \: that \\ \\ \sf \to \: (x - x_{1}) {}^{2} + (y - y_{1}) = (x - x_{2}) {}^{2} + (y - y_{2}) {}^{2}$

$\sf \to \: (h - 0) {}^{2} + (k - 1) {}^{2} = (h - 2) {}^{2} + (k - 4) {}^{2} \\ \\ \sf \to \: { \cancel{h}^{2}} + { \cancel{k}^{2}} + 1 - 2k = { \cancel{h}^{2}} + 4 - 4h + { \cancel{k}^{2}} + 16 - 8k \\ \\ \sf \to \: 1 - 2k = 4 - 4h + 16 - 8k \\ \\ \sf \to \: 4h + 6k - 19 = 0 \: \: \: .......(1)$

$\sf \to \: as \: we \: know \: that \: \\ \\ \sf \to \: slope \: of \: oa \: = \frac{k - 4}{h - 2} \\ \\ \sf \to \: slope \: of \: tangent \: at \: point \: (2 ,4) \: at \: y \: = {x}^{2} = 4 \\ \\ \sf \to \: m_{1} \times m_{2} = - 1 \\ \\ \sf \to \:( slope \: of \: oa) \times (slope \: of \: tangent \: at \: a) = - 1 \\ \\ \sf \to \frac{k - 4}{h - 2} \times 4 = - 1$

$\sf \to \: \dfrac{4k - 16}{h - 2} = - 1 \\ \\ \sf \to \: 4k - 16 = - h + 2 \\ \\ \sf \to \: 4k + h = 18 \: \: \: .......(2)$

$\sf \to \: from \: equation \: (1) \: \: \: and \: \: \: (2) \: \: we \: get \\ \\ \sf \to \: h \: = 18 - 4k \: \: .....(3) \\ \\ \sf \to \: put \: the \: value \: in \: equation \: (1) \\ \\ \sf \to \: 4(18 - 4k) + 6k - 19 = 0 \\ \\ \sf \to \: 72 - 16k + 6k - 19 = 0 \\ \\ \sf \to \: - 10k = - 53 \\ \\ \sf \to \: k \: = \frac{53}{10}$

$\sf \to \: put \: the \: value \: in \: equation \: (3) \: we \: get \\ \\ \sf \to \: h \: = 18 - 4 \times \frac{53}{10} \\ \\ \sf \to \: h \: = 18 - \frac{106}{5} \\ \\ \sf \to \: h \: = \frac{90 - 106}{5} \\ \\ \sf \to \: h \: = \frac{ - 16}{5}$

$\sf \to \: coordinate \: of \: centre \: = ( \dfrac{ - 16}{5}, \dfrac{53}{10} )$