# Find the equation of a circle passing through (3,7) and tangent to the lines x-3y+8=0 and y=3x

1
by casiminagibbson

2015-05-15T18:57:51+08:00

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Lines tangent:
x-3y+8=0
y=3x

Simply substitute the values:
x-3(3x)+8=0
x-9x+8=0
-8x+8=0
Subtract both sides by 8
-8x=-8
Divide both sides by -8
x=1

Substitute the value of x:
y=3(1)
y=3

The coordinate is: (1,3) which is where it is tangent..

Both of them are endpoints of the circle so I recommend you draw a line connecting both of them, then attach, then the midpoint of it should be the center

M:((1+3)/2 , (3+7)/2)
M:(4/2 , 10/2)
M:(2,5)

Find the distance from the radius to one of its endpoints..
d=√[(x2-x1)²+(y2-y1)²]
d=√[(3-2)²+(7-5)²]
d=√[1+4]
d=√5

The center is (2,5) Now 2 is the h and 5 is the k from the equation:
(x-h)²+(y-k)²=r²
(x-2)²+(y-5)²=(√5)²
x²-4x+4+y²-10y+25=5
x²-4x+y²-10y+24=0

Hope this helps =)