Question 707202
Write the standard and general equation of a circle through (2,1) and (3,5) and having its center on the line 8x + 5y = 8.

<pre>
There is more than one way to do this, but let's do it the
way the tutor above said to do it:

The standard equation of a circle is

(x-h)² + (y-k)² = r²

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(2,1,"(2,1)"), locate(3,5,"(3,5)"),
circle(2,1,.1), circle(3,5,.1), line(-19,32,16,-24) )}}}

The line connecting (2,1) amd (3,5) is a chord of the circle. 
{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(2,1,"(2,1)"), locate(3,5,"(3,5)"),green(line(2,1,3,5)),
circle(2,1,.1), circle(3,5,.1), line(-19,32,16,-24) )}}}

So we can use the fact that the perpendicular bisector of a chord 
passes through the center of the circle. That's the red line below

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(2,1,"(2,1)"), locate(3,5,"(3,5)"),green(line(2,1,3,5)),red(line(-15,7.4,13,.3)),
circle(2,1,.1), circle(3,5,.1), line(-19,32,16,-24) )}}}

We get the slope of the green chord using the slope formula:

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}
m = {{{(5-1)/(3-2)}}}
m = {{{4/1}}}
m = 4

So the slope of the red perpendicular bisector of the green
chord is it's "negative reciprocal" or {{{-1/4}}}.

We get the midpoint of the green chord using the midpoint
formula:

Midpoint = {{{(matrix(1,3,(x[1]+x[2])/2, ",",(y[1]+y[2])/2))}}}

Midpoint = {{{(matrix(1,3,(2+3)/2, ",",(1+5)/2))}}}

Midpoint = {{{(matrix(1,3,5/2, ",",6/2))}}} 

Midpoint = {{{(matrix(1,3,5/2, ",",3))}}}

So the red perpendicular bisector of the chord is the line with
slope {{{-1/4}}} that goes through the point {{{(matrix(1,3,5/2, ",",3))}}}

We use the point slope formula:

y - y<sub>1</sub> = m(x - x<sub>1</sub>)

y - 3 = {{{-1/4}}}(x - 5/2)

y - 3 = {{{-1/4}}}x + 5/8

Multiply through by 8

8y - 24 = -2x + 5
2x + 8y = 29

So to find the center we solve the system

{{{system(8x + 5y = 8,2x + 8y = 29)}}}

and get the point (x,y) = ({{{-3/2}}},4)

That's the center (h,k) = ({{{-3/2}}},4)

So we can draw in the circle:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(2,1,"(2,1)"), locate(3,5,"(3,5)"),green(line(2,1,3,5)),red(line(-15,7.4,13,.3)),circle(-3/2,4,sqrt(85)/2),
circle(2,1,.1), circle(3,5,.1), line(-19,32,16,-24) )}}}

So the equation of the circle, so far is

(x-({{{-3/2}}})² + (y-4)² = r²

(x+{{{3/2}}})² + (y-4)² = r²

And since we know it goes through (2,1) we can
substitute that point:

(2+{{{3/2}}})² + (1-4)² = r²
({{{4/2}}}+{{{3/2}}})² + (-3)² = r²
({{{7/2}}})² + 9 = r²
{{{49/4}}} + 9 = r²
{{{49/4}}} + {{{36/4}}} = r²
{{{85/4}}} = r²

So the standard equation of the circle is

(x+{{{3/2}}})² + (y-4)² = {{{85/4}}}

To find the general equatrion of the circle,
nultiply that out:

x² + 3x + {{{9/4}}} + y² - 8y + 16 = {{{85/4}}}

Multiply through by 4

4x² + 12x + 9 + 4y² - 32y + 64 = 85

4x² + 12x + 73 + 4y² - 32y = 85

4x² + 4y² + 12x - 32y - 12 = 0

Divide through by 4

x² + y² + 3x - 8y - 3 = 0

That's the general equation of the circle.

Edwin</pre>