Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

Find the equation of the circle in general forn if the diameter is 12 and the center is 8,1? Thabk you for the answer

2 Answers

Jun 20, 2018

${(x - 8)}^{2} + {(y - 1)}^{2} = 6^{2}$ $(x-8)^2+(y-1)^2=6^2$

Explanation:

If the diameter is $12$ $12$ then the radius is $\frac{12}{2} = 6$ $12/2=6$

The general equation of a circle with center $(a, b)$ $(a,b)$ and radius $r$ $r$ is
$XXX {(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $color(white)("XXX")(x-a)^2+(y-b)^2=r^2$

Jun 25, 2018

${(x - 8)}^{2} + {(y - 1)}^{2} = 6^{2}$ $(x-8)^2+(y-1)^2=6^2$

Explanation:

The equation of a circle is basically using Pythagoras $a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$

It is just that we are using the x-axis value as say $a$ $a$ , the y-axis value as say $b$ $b$ and the radius of the circle as $c$ $c$

If the centre is at the origin we have $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

This works very well when centred at at the origin but what if it is elsewhere?

We 'hypothetically' take it back to the origin

So if the centre is at 8 on the x-axis we write $(x - 8)$ $(x-8)$
So if the centre is at 1 on the y-axis we write $(y - 1)$ $(y-1)$

The length of the radius is unchanged giving:

${(x - 8)}^{2} + {(y - 1)}^{2} = r^{2} = {(\frac{12}{2})}^{2}$ $(x-8)^2+(y-1)^2=r^2=(12/2)^2$

Related questions

Impact of this question

3087 views around the world

You can reuse this answer
Creative Commons License