The equation of a circle is $x^2+y^2+8x−14y+56=0$ . What is the radius of the circle?

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Answer 1 · 2018-01-04T06:15:29

r = 3

Equation General:

$(x - h)^2 + (y - k)^2 = r^2$

Re-arranging:

$(x^2 + 8x) + (y^2 - 14y) + 56= 0$

Making squares by analogy (method)
1): divide each no square variable coefficient
$8/2 = 4$
$14/2 = 7$

2): make square each divided coefficient
$4^2 => 16$
$7^2 => 49$

3) Substitute previous coefficients in general equation:
$(x^2 + 8x + 16) + (y^2 - 14y + 49) + (56 - 16 - 49)= 0$

$(x + 4)^2 + (y - 7)^2 = (9)^2$

$r^2 = 9$

$r = 3$

The equation of a circle is x^2+y^2+8x−14y+56=0x^2+y^2+8x−14y+56=0 . What is the radius of the circle?