a circle touches y axis at (0,5) and whose centre lies on the line 2x+y=13,find the equation of the circle
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perpendicular distance of Pt. from line
r=2(0)+5-13/√5=-8/√5
r=8/√5
..radius
for centre let the X Co ordinate of point =h
therefore y Co ordinate =13-2h
Now centre=(h, 13-2h)
we know radius
therefore apply distance formula
√(h-0)^2 + (13-2h-5)^2=(8/√5)
solve for h and put in point of centre.. now u have centre and radius both.... and we know if (m, n) is centre and r is radius then eqn of circle is (x-m) ^2+(y-n)^2=r^2
r=2(0)+5-13/√5=-8/√5
r=8/√5
..radius
for centre let the X Co ordinate of point =h
therefore y Co ordinate =13-2h
Now centre=(h, 13-2h)
we know radius
therefore apply distance formula
√(h-0)^2 + (13-2h-5)^2=(8/√5)
solve for h and put in point of centre.. now u have centre and radius both.... and we know if (m, n) is centre and r is radius then eqn of circle is (x-m) ^2+(y-n)^2=r^2
dragomegaman:
from where did root 5 come?
under root of sum of squares of coefficient of X and y..... I. e. ,(√2^2+1)=√5
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