# Find the equation of the circle with diameter AB where A and B are the points (-1,2) and (3,3) respectively?

Sep 15, 2015

The answer could be find out by knowing some formulas of coordinate geometry.
1) standard eqn.= (x−a)^2 + (y−b)^2 = r^2 (for this question i,e, radius)
2) midpoint= $\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}$
3) distance= sqrt${\left({x}_{2} - {x}_{1}\right)}^{2}$+${\left({y}_{2} - {y}_{1}\right)}^{2}$

#### Explanation:

we are given , A(-1,2) & B(3,3)
here ${x}_{1}$= -1
${y}_{1}$= 2
${x}_{2}$= 3
${y}_{2}$= 3

Therefore, AB= sqrt${\left({x}_{2} - {x}_{1}\right)}^{2}$+${\left({y}_{2} - {y}_{1}\right)}^{2}$

= sqrt${\left(3 - \left(- 1\right)\right)}^{2}$+${\left(3 - 2\right)}^{2}$

sqrt${\left(4\right)}^{2}$+${\left(1\right)}^{2}$

sqrt$\left(16\right)$+$\left(1\right)$

sqrt$17$.....................(Diameter)

so radius = 1/2 * diameter
= sqrt$17$/2

now the radius can also be got by calculating the midpoint
so the next step is:
wkt,midpoint is given by =$\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}$

then,we get $\frac{- 1 + 3}{2}$ , $\frac{2 + 3}{2}$

= $\left(\frac{2}{2}\right)$ , $\left(\frac{5}{2}\right)$

= (1 , 2.5)
Hence we get the values of a and b respectively.
Putting the values we get,
(x−a)^2 + (y−b)^2 = r^2

= (x−1)^2 + (y−2.5)^2 = (sqrt17/2)^2
= (x−1)^2 + (y−2.5)^2 = $\left(\frac{17}{4}\right)$
= (x−1)^2 + (y−2.5)^2 = 4.25
= (x−1)^2 + (y−2.5)^2 = 4.25........................(answer)

Sep 17, 2015

${x}^{2} + {y}^{2} - 2 x - 5 y + 3 = 0$

#### Explanation:

Firstly, we can find the centre of the circle by finding the midpoint of $A B$. Since it the midpoint or the centre $\left(h , k\right)$ cut AB with equal ratio;

$C e n t r e \left(h , k\right) = \left(\frac{- 1 + 3}{2} , \frac{2 + 3}{2}\right)$

$\left(h , k\right) = \left(1 , \frac{5}{2}\right)$

Then we can find the radius , $r$ of the circle by using the equation;

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

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

Substitute the coordinate $\left(h , k\right) = \left(1 , \frac{5}{2}\right)$ and any of $A$ or $B$ coordinates into equation. In this calculation I choose $B$.

r^=sqrt((3-1)^2+(3-5/2)^2

$r = \sqrt{\frac{17}{4}}$=$\frac{\sqrt{17}}{2}$

To find $c$, we can use the equation $c = {h}^{2} + {k}^{2} - {r}^{2}$ and substitute $\left(h , k\right) = \left(1 , \frac{5}{2}\right)$ and $r = \frac{\sqrt{17}}{2}$ into it.

$c = {\left(1\right)}^{2} + {\left(\frac{5}{2}\right)}^{2} - {\left(\frac{\sqrt{17}}{2}\right)}^{2}$

$c = 3$

Then we know that the equation of circle is known as;

${x}^{2} + {y}^{2} - 2 h x - 2 k y + c = 0$

Substitute only $\left(h , k\right) = \left(1 , \frac{5}{2}\right)$ and $c = 3$ and we get;

${x}^{2} + {y}^{2} - 2 \left(1\right) x - 2 \left(\frac{5}{2}\right) y + 3 = 0$

${x}^{2} + {y}^{2} - 2 x - 5 y + 3 = 0$