Question bank/JEE Main/Mathematics

If $C_{1}:x^{2} + y^{2} = (3 + 2\sqrt{2})^{2}$is a circle and PA
and PB are a pair of tangents on C_1,where P is any point on the
director circle of C_1 then the radius of the smaller circle which
touches C_1 externally and also the two tangents PA and PB is;fn $C_{1}:x^{2} + y^{2} = (3 + 2\sqrt{2})^{2}$,d o`Ùk gS rFkk PA vkSj
PB ij tgk¡ P, C_1 o`Ùk ij dksbZ
fcUnq gS rc NksVs o`Ùk dh f=T;k gksxh tks C_1 dks cká rFkk nks PA rFkk PB

A$2\sqrt{3} - 3$B$2\sqrt{2} - 1$C1D$2\sqrt{2} - 2$

Show answer and solution▾

$AQ = 3 + 2\sqrt{2}$
$PQ = 3\sqrt{2} + 4$
Let r be the required radius
Then rc
$3\sqrt{2} + 4 = 3 + 2\sqrt{2} + r + r\sqrt{2}$
r = 1

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx