What are the coordinates of the center of the circle shown below? x^2+y^2-12x-6y+9=0
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<u>ΔACB</u> <u>ΔCDA</u>
AC² + BC² = AB² AD² + CD² = AC²
BC² = AB² - AC² BC² + CD² = AC² (AD=BC is given)
BC² = AC² - CD²
AB² - AC² = AC² - CD² (both sides were = to BC²)
AB² + CD² = 2AC²
(3)² + (√2)² = 2AC² (AB=3 and CD=√2 were given)
9 + 2 = 2AC²
11 = 2AC²
= AC²
= AC
= AC