Jawabanmu

2013-12-19T07:51:07+07:00
Panjang diameter =  panjang PQ =  \sqrt{(3-0)^2+(0-(-6))^2}= \sqrt{9+36}  = \sqrt{45}

maka jari-jari =  \frac{1}{2}.d= \frac{1}{2} \sqrt{45}

pusat = titik tengah PQ = ( \frac{3+0}{2} \ , \  \frac{0+(-6)}{2})=( \frac{3}{2}\ ,\-3 )

pers lingkaran
(x-a)^2+(y-b)^2=r^2
(x- \frac{3}{2} )^2+(y-(-3))^2=( \frac{1}{2} \sqrt{45} ) ^2

x^2-3x+ \frac{9}{4}+y^2+6y+9= \frac{1}{4} (45)

x^2+y^2-3x+6y+ \frac{9}{4}+9- \frac{45}{4}=0

x^2+y^2-3x+6y+ \frac{9+36-45}{4}=0

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