The equation of the circle whose centre lies on the line x-4y=1 and which passes through the points (3 ,7 ) and (5 ,5 ) is

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Correct answer: A. (a) : Consider the line first, we have x-4y=1 ⇒x=1+4y ∴ Any point on the line is of the form (4a+1 ,a ), which included the centre of circle also. This point is equidistant from any point on the circumference of the circle. ∴ (4a+1-3 )^2+ (a-7 )^2= (4a+1-5 )^2+ (a-5 )^2 ⇒ 16a^2-16a+4+a^2-14a+49=16a^2-32a+16 +a^2-10a+25 ⇒ -12a=12 ⇒a=-1 ∴ Centre of circle= (-4+1 ,-1 )= (-3 ,-1 ) Now , radius (r )= (-3-3 )^2+ (-1-7 )^2= 100=10 ∴ Equation of circle is (x+3 )^2+ (y+1 )^2=10^2 i.e., x^2+y^2+6x+2y-90=0

$The equation of the circle whose centre lies on the line x - 4 y = 1 and which passes through the points (3, 7) and (5, 5) is$

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