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JEE MainMathematicsToughCircles

AB is any chord of the circle x2+y2−6x−8y−11=0x^{2} + y^{2} - 6x - 8y - 11 = 0
which subtends an angle
π2\frac{\pi}{2}at (1,2)(1,2). If locus of midpoint
of AB is a circle
x2+y2−2ax−2by−c=0x^{2} + y^{2} - 2ax - 2by - c = 0 the value of
(a+2b+3c)(a + 2b + 3c) is
o`r
x2+y2−6x−8y−11=0x^{2} + y^{2} - 6x - 8y - 11 = 0 sbZ thok AB tks fcUnq
(1, 2)(1,\ 2)ijπ2\frac{\pi}{2}dks.k cukrh AB fcUnq dk
fcUnqiFk,d o`r
x2+y2−2ax−2by−c=0x^{2} + y^{2} - 2ax - 2by - c = 0gS rc
(a+2b+3c)(a + 2b + 3c)dk eku

Show answer and solution
PQ=PAPQ = PA
(h−1)2+(k−2)2=36−(h−3)2−(k−4)2(h - 1)^{2} + (k - 2)^{2} = 36 - (h - 3)^{2} - (k - 4)^{2}
x2+y2−4x−6y−3=0x^{2} + y^{2} - 4x - 6y - 3 = 0

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

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