Question

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

• A. $x^2+y^2+4x-4y+3=0$
• B. $3x^2+3y^2-4x+6y=0$
• C. $3x^2+3y^2+4x-4y=0$
• D. $3x^2+3y^2-4x+4y=0$

Answer 1

The correct option is B $3x^2+3y^2-4x+6y=0$

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

$(x-a{)}^2+(y-b{)}^2=r^2$

plug in the points

$solve3equationsfor3unknowns\left(abandr\right)$

plug those back into the eq of a circle

$(0-a{)}^2+(0-b{)}^2=r^2$

$a^2+b^2=r^2$

$(0-a{)}^2+(2-b{)}^2=r^2$

$a^2+b^2-4b-4=r^2$ $[since{a}^2+b^2=r^2,-4b-4=0,orb=-1]$

$(3-a{)}^2+(0-b{)}^2=r^2$

$9-6a+a^2+b^2=r^2$ $[againsince{a}^2+b^2=r^2,9-6a=0ora=2/3]$

$solvefor{r}^2$

$(2/3{)}^2+(-1{)}^2=r^2$

$13/9=r^2$

$socircleeqis(x-a{)}^2+(y-b{)}^2=r^2$

$(x-2/3{)}^2+(y+1{)}^2=\frac{13}{9}$

$x^2-\frac{4}{3}x+4/9+y^2+2y+1=\frac{13}{9}$

$x^2-\frac{4}{3}x+y^2+2y=0$

$3x^2-4x+3y^2+6y=0$