Tentukan pusat jari jari lingkaran dengan persamaan a. x²+y²+8x-6y+17=0.... Question from @jeyes

September 2018 2 19 Report

Tentukan pusat jari jari lingkaran dengan persamaan a. x²+y²+8x-6y+17=0

vyanistyanggara Dik: A=8, B=-6, C=17
p= (-A/2 , -B/2)
=(-8/2 , -(-6)/2)
=(-4,3)
r= √A²/4 + B²/4 - C
=√8²/4 + (-6)²/4 - 17
=√16 + 9 - 17
=√8
=2√2

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IcukSugiarto $\bold{\underline{Solution:}}\\\\rumus\ umum:\\\boxed{x^{2}+y^{2}+Ax+By+C=0}\\\\sehingga\ :\\A=8~~~;B=-6~~~;C=17\\\\maka\ untuk\ pusat\ lingkaran:\\p= (\frac{-A}{2},\frac{-B}{2})\\p= (\frac{-8}{2},\frac{-(-6)}{2})\\p=(-4,3)$

$untuk\ jari\ jari\ lingkaran:\\\boxed{r= \sqrt{ \frac{A^{2}}{4}+\frac{B^{2}}{4} -C} }\\\boxed{r= \sqrt{ \frac{8^{2}}{4}+\frac{(-6)^{2}}{4} -17} }\\\boxed{r= \sqrt{ \frac{64}{4}+\frac{36}{4} -17} }\\\boxed{r= \sqrt{ \frac{100}{4} -17}}\\\boxed{r= \sqrt{ 25-17}}\\\boxed{r= \sqrt{ 8}}\\\boxed{r= \sqrt{4 \times 2}}\\\boxed{r= 2\sqrt{ 2}}$

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