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search-thumbnail-Lenter (1,-/); radius 8 
C. 
$1$ $4x^{2}+4y^{2}+12x+-4y-90=$ $0$ 
$4x^{2}+4y2+12x-4y=90$ 
Divide both $sidosby4$ 
$\dfrac {4x^{2}+4y^{2}+12x-4y=} {4}\dfrac {90} {4}$ 
$x^{2}+4y^{2}+12x-4y=\dfrac {90} {4}$ 
$2$ $x^{2}+y^{2}+8x-6y-39=0$ $-$ 
3. $x^{2}+y^{2}-10x+16y-32=0$
10th-13th grade
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