# How do you graph #8x ^ { 2} + 72x + 8y ^ { 2} + 16y - 342= 0#?

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Shwetank Mauria
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Mar 9, 2018

#### Answer:

It is equation of a circle with center

#### Explanation:

Observe that

#8x^2+72x+8y^2+16y-342=0# is a quadratic equation,- in which coefficients of
#x^2# and#y^2# are equal and - their is no term containing
#xy# .

Hence, it represents te equation of a circle. Let us convert it to appropriate form. Diving each term by

or

i.e.

Hence, this is equation of a point whose distance from

It is equation of a circle with center

graph{(8x^2+72x+8y^2+16y-342)((x+9/2)^2+(y+1)^2-0.08)=0 [-20, 20, -10, 10]}

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