Find the equation of circle with radius 4 units and whose centre lies on the line 13x+4y=32 and which touches the line 3x+4y+28=0.?

1 Answer
Nov 26, 2017

see explanation.

Explanation:

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Let equation of L1 be 13x+4y=32, and equation of L2 be 3x+4y+28=0.
Let center of the circle be (h,k), as shown in the figure.
given that (h,k) lies on L1,
Substituting h=x, y=k into 13x+4y=32, we get,
13h+4k=32,
=> k=(32-13h)/4
recall that the perpendicular distance from a point P(h,k) to a line ax+by+c=0 is d=|ah+bk+c|/sqrt(a^2+b^2),
given that r=4, and the circle touches L2,
=> 4=|3h+4xx(32-13h)/4+28|/(sqrt(3^2+4^2))
=> 4=|3h+32-13h+28|/5
=> 20=|-10h+60|
=> h=4 or 8
when h=4, k=(32-13xx4)/4=-5
when h=8, k=(32-13xx8)/4=-18
=> (h,k)=(4,-5), (8,-18)
equations of the circles are:
(x-4)^2+(y+5)^2=16,
and (x-8)^2+(y+18)^2=16