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Question

Find the equation of the circle which circumscribes the triangle formed by the lines x=0,y=0 and 2x+3y=6.

A
x2+y2+4x4y+3=0
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B
3x2+3y24x+6y=0
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C
3x2+3y2+4x4y=0
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D
3x2+3y24x+4y=0
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Solution

The correct option is B 3x2+3y24x+6y=0

Since the circle circumscribes the triangle, the three corners of the triangle are points on the circle. Find the points of intersection of the three lines.

(0,0),(0,2),(3,0)

eq of a circle is

(xa)2+(yb)2=r2

plug in the points

solve 3 equations for 3 unknowns (a b and r)

plug those back into the eq of a circle

(0a)2+(0b)2=r2

a2+b2=r2


(0a)2+(2b)2=r2

a2+b24b4=r2 [since a2+b2=r2, 4b4=0, or b=1]


(3a)2+(0b)2=r2

96a+a2+b2=r2 [again since a2+b2=r2, 96a=0 or a=2/3]

solve for r2

(2/3)2+(1)2=r2

13/9=r2

so circle eq is (xa)2+(yb)2=r2

(x2/3)2+(y+1)2=139

x243x+4/9+y2+2y+1=139

x243x+y2+2y=0

3x24x+3y2+6y=0


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