Question

A triangle has corners at `(9 ,4 )`, `(7 ,5 )`, and `(3 ,6 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Reqd. Area=`212.5pi sq.unit~=667.25sq.unit.`

Explanation:

To find the reqd. area of the circumcircle of the `Delta ABC`, where `A(9,4), B(7,5), C(3,6)` , we have to first find out the radius of the circle, say `R.`

Suppose that, pt.`P(x,y)` is the circumcentre of `Delta ABC.`

Then, dist. `PA`=dist. `PB`=dist. `PC,` each `=R.`

`:. (PA)^2=(PB)^2=(PC)^2.` Using Dist. Formula, we get,

`(x-9)^2+(y-4)^2=(x-7)^2+(y-5)^2=(x-3)^2+(y-6)^2.`

`:. -18x+81-8y+16=-14x+49-10y+25=-6x+9-12y+36`

`:. -4x+2y+23=0..........(1),` [using first & second eqns.], &
`-8x+2y+29=0.............(2)`,[using second & third eqns.]

Then, `(1)-(2)` gives, `4x-6=0,` or, `x=3/2`, then by `(1), y=-17/2`

So, the circumcentre of `Delta ABC` is `P(3/2,-17/2).`

Hence, `R^2=(CP)^2=(3-3/2)^2+(6+17/2)^2=9/4+841/4=850/4=425/2,` giving the Area of Circumcircle of `DeltaABC`=`pi*R^2=425/2pi=212.5pi~=667.25`