What is the area of the circle?

[Math Tutor]

What is the area of the circle?

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[Guest]

85[tex]\pi[/tex]

[Guest]

Also can be done by 3 points A(0,0), B(16,4) and C(16,8). Now if center is Q(a,b), then radius AQ=BQ=CQ..
EQ1 [tex]AQ^{2} = (a-0)^{2}+(b-0)^{2} = a^{2}+b^{2}[/tex]
EQ2 [tex]BQ^{2} = (16-a)^{2}+(4-b)^{2} = 272 - 32a - 8b +a^{2}+b^{2}[/tex]
EQ3 [tex]CQ^{2} = (16-a)^{2}+(8-b)^{2} = 320 - 32a - 16b +a^{2}+b^{2}[/tex]
Solving these
EQ1 = EQ3
[tex]a^{2}+b^{2} = 272 - 32a - 16b +a^{2}+b^{2}[/tex]
[tex]32a + 16b = 320[/tex]
[tex]2a + b = 20[/tex]

EQ2 = EQ3
[tex]272 - 32a - 8b +a^{2}+b^{2} = 320 - 32a - 16b +a^{2}+b^{2}[/tex]
[tex]16b - 8b = 320 - 272[/tex]
[tex]8b = 48[/tex]
[tex]b=6[/tex]
and
[tex]a=7[/tex]

So
[tex]AQ^{2} = a^{2}+b^{2} = 7^{2}+6^{2} = 85[/tex]
and Area is [tex]\pi85[/tex]

[Guest]

Also can be done by 3 points A(0,0), B(16,4) and C(16,8). Now if center is Q(a,b), then radius AQ=BQ=CQ..
EQ1 [tex]AQ^{2} = (a-0)^{2}+(b-0)^{2} = a^{2}+b^{2}[/tex]
EQ2 [tex]BQ^{2} = (16-a)^{2}+(4-b)^{2} = 272 - 32a - 8b +a^{2}+b^{2}[/tex]
EQ3 [tex]CQ^{2} = (16-a)^{2}+(8-b)^{2} = 320 - 32a - 16b +a^{2}+b^{2}[/tex]
Solving these
EQ1 = EQ3
[tex]a^{2}+b^{2} = 320 - 32a - 16b +a^{2}+b^{2}[/tex]
[tex]32a + 16b = 320[/tex]
[tex]2a + b = 20[/tex]

EQ2 = EQ3
[tex]272 - 32a - 8b +a^{2}+b^{2} = 320 - 32a - 16b +a^{2}+b^{2}[/tex]
[tex]16b - 8b = 320 - 272[/tex]
[tex]8b = 48[/tex]
[tex]b=6[/tex]
and
[tex]a=7[/tex]

So
[tex]AQ^{2} = a^{2}+b^{2} = 7^{2}+6^{2} = 85[/tex]
and Area is [tex]\pi85[/tex]