# What is the standard form of the equation of a circle with centre is at point (5,8) and which passes through the point (2,5)?

Jan 5, 2016

${\left(x - 5\right)}^{2} + {\left(y - 8\right)}^{2} = 18$

#### Explanation:

standard form of a circle is ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

where (a , b ) is the centre of the circle and r = radius.

in this question the centre is known but r is not. To find r , however ,

the distance from the centre to the point ( 2 , 5 ) is the radius . Using

the distance formula will allow us to find in fact ${r}^{2}$

${r}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$

now using (2 , 5 ) = $\left({x}_{2} , {y}_{2}\right) \mathmr{and} \left(5 , 8\right) = \left({x}_{1} , {y}_{1}\right)$

then ${\left(5 - 2\right)}^{2} + {\left(8 - 5\right)}^{2} = {3}^{2} + {3}^{2} = 9 + 9 = 18$

equation of circle : ${\left(x - 5\right)}^{2} + {\left(y - 8\right)}^{2} = 18$
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Jan 5, 2016

I found: ${x}^{2} + {y}^{2} - 10 x - 16 y + 71 = 0$

#### Explanation:

The distance $d$ between the centre and the given point will be the radius $r$.
We can evaluate it using:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
So:
$r = d = \sqrt{{\left(2 - 5\right)}^{2} + {\left(5 - 8\right)}^{2}} = \sqrt{9 + 9} = 3 \sqrt{2}$
Now you can use the general form of the equation of a circle with centre at $\left(h , k\right)$ and radius $r$:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
And:
${\left(x - 5\right)}^{2} + {\left(y - 8\right)}^{2} = {\left(3 \sqrt{2}\right)}^{2}$
${x}^{2} - 10 x + 25 + {y}^{2} - 16 y + 64 = 18$
${x}^{2} + {y}^{2} - 10 x - 16 y + 71 = 0$