Question 150278
Find the equation of the circle described each. 
a)The circle is tangent to both coordinate axes and contains the point (6,3). 

<pre><font size = 4 color = "indigo"><b>

First plot the point (6,3)

{{{drawing(700,700,-5,35,-5,35, graph(700,700,-5,35,-5,35), locate(6-.15,3+.4,"@"), locate(6+.5,3+.5,"(6,3)") )}}}

I can see how there could be two different solutions, by drawing in
these:

{{{drawing(700,700,-5,35,-5,35, graph(700,700,-5,35,-5,35), locate(6-.15,3+.4,"@"), locate(6+.5,3+.5,"(6,3)"),
circle(3,3,3),circle(15,15,15) )}}}

The equation of a circle with center (h,k) and radius r is

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

Draw in radii to the axes:

{{{drawing(700,700,-5,35,-5,35, graph(700,700,-5,35,-5,35), locate(6-.15,3+.4,"@"), locate(6+.5,3+.5,"(6,3)"), rectangle(0,0,3,3),rectangle(0,0,15,15),
circle(3,3,3),circle(15,15,15) )}}}

We can see that since the circle has to be tangent to both axes, 
its center has to have the same x and y coordinates. and also that
the radius has to be equal to h as well.  So we can see that all
three values h, k, and r, must all be the same. So let them all be 
h, i.e., h = k = r, and we have

{{{drawing(700,700,-5,35,-5,35, graph(700,700,-5,35,-5,35), locate(6-.15,3+.4,"@"), locate(6+.5,3+.5,"(6,3)"), rectangle(0,0,3,3),rectangle(0,0,15,15), locate(3,3,"(h,h)"),locate(15,15,"(h,h)"),
circle(3,3,3),circle(15,15,15) )}}}

So since {{{h=k=r}}}, we have

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

Now since it contains the point (6,3) we can substitute that in

{{{(6 - h)^2 + (3-h)^2 = h^2}}}

{{{(6-h)(6-h) + (3-h)(3-h) = h*2}}}

{{{(36-6h-6h+h^2)+(9-3h-3h+h^2) = h^2}}}

{{{(36-12h+h^2)+(9-6h+h^2) = h^2}}}

{{{36-12h+h^2+9-6h+h^2 = h^2}}}

That simplifies to

{{{h^2-18h+45=0}}}

Factoring:

{{{(h-3)(h-15)=0}}}

{{{h-3=0}}} or {{{h=3}}}
{{{h-15=0}}} or {{{h=15}}}

So we two values of {{{h}}}, so

the two circles' equations are

{{{(x - 3)^2 + (y-3)^2 = 3^2}}} and {{{(x - 15)^2 + (y-15)^2 = 15^2}}}

{{{(x - 3)^2 + (y-3)^2 = 9}}} and {{{(x - 15)^2 + (y-15)^2 = 225}}}



b)The circle is circumscribed about the triangle whose vertices are (-1,-3),(-2,4),and (2,1). 

We plot the points:

{{{drawing(400,400,-10,10,-10,10, locate(-2-.2,4.4,"@"),graph(400,400,-10,10,-10,10), locate(2-.2,1.4,"@"), locate(-1-.2,-3+.4,"@") )}}}

We draw the triangle:

{{{drawing(400,400,-10,10,-10,10, locate(-2-.2,4.4,"@"), graph(400,400,-10,10,-10,10),
locate(2-.2,1.4,"@"), locate(-1-.2,-3+.4,"@"), triangle(-1,-3,-2,4,2,1) 
 )}}}

The equation of a circle with center ({{{h}}},{{{k}}}) and radius {{{r}}} is:

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

We substitute the point ({{{x}}},{{{y}}}) = ({{{-1}}},{{{-3}}}) 

{{{(-1-h)^2+(-3-k)^2=r^2}}}

{{{(-1-h)(-1-h)+(-3-k)(-3-k)=r^2}}}

{{{(1+h+h+h^2)+(9+3k+3k+k^2)=r^2}}}

{{{(1+2h+h^2)+(9+6k+k^2)=r^2}}}

{{{1+2h+h^2+9+6k+k^2=r^2}}}

{{{10+2h+h^2+6k+k^2=r^2}}}

Get all squared terms on right:

{{{10+2h+6k=r^2-h^2-k^2}}}

We substitute the point ({{{x}}},{{{y}}}) = ({{{-2}}},{{{4}}}) 

{{{(-2-h)^2+(4-k)^2=r^2}}}

{{{(-2-h)(-2-h)+(4-k)(4-k)=r^2}}}

{{{(4+2h+2h+h^2)+(16-4k-4k+k^2)=r^2}}}

{{{(4+4h+h^2)+(16-8k+k^2)=r^2}}}

{{{4+4h+h^2+16-8k+k^2=r^2}}}

{{{20+4h+h^2-8k+k^2=r^2}}}

Get all squared terms on right:

{{{20+4h-8k=r^2-h^2-k^2}}}

We substitute the point ({{{x}}},{{{y}}}) = ({{{2}}},{{{1}}}) 

{{{(2-h)^2+(1-k)^2=r^2}}}

{{{(2-h)(2-h)+(1-k)(1-k)=r^2}}}

{{{(4-2h-2h+h^2)+(1-k-k+k^2)=r^2}}}

{{{(4-4h+h^2)+(1-2k+k^2)=r^2}}}

{{{4-4h+h^2+1-2k+k^2=r^2}}}

{{{5-4h+h^2-2k+k^2=r^2}}}

Get all squared terms on right:

{{{5-4h-2k=r^2-h^2-k^2}}}

So we have the three equations:


{{{10+2h+6k=r^2-h^2-k^2}}}
{{{20+4h-8k=r^2-h^2-k^2}}}
 {{{5-4h-2k=r^2-h^2-k^2}}}

Since the right sides of all three equations are equal,
then so are the left sides:

{{{10+2h+6k=20+4h-8k=5-4h-2k}}}

Using the first two

{{{10+2h+6k=20+4h-8k}}}

{{{-2h+14k=10}}}

Using the first and third:

{{{10+2h+6k=5-4h-2k}}}

{{{6h+8k=-5}}}

So we solve these two equations by 
substitution of elimination:

{{{-2h+14k=10}}}
{{{6h+8k=-5}}}

and get (h,k) = ({{{-3/2}}},{{{1/2}}})

To find r we go back to

{{{10+2h+6k=r^2-h^2-k^2}}}

{{{10+2(-3/2)+6(1/2)=r^2-(-3/2)^2-(1/2)^2}}}

{{{10-3+3=r^2-(9/4)-(1/4)}}}

{{{10=r^2-10/4}}}

{{{10=r^2-5/2}}}

{{{20=2r^2-5}}}

{{{25=2r^2}}}

{{{25/2=r^2}}}

{{{sqrt(25/2)= r}}}

Therefore the equation 

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

becomes

{{{(x-(-1/2))^2+(y-3/2)^2=(sqrt(25/2))^2}}}

{{{(x+1/2)^2+(y-3/2)^2=25/2}}}

So we plot the center ({{{-3/2}}},{{{1/2}}})

{{{drawing(400,400,-10,10,-10,10, locate(-2-.2,4.4,"@"), graph(400,400,-10,10,-10,10), locate(-3/2-.2,1/2+.4,"@"),
locate(2-.2,1.4,"@"), locate(-1-.2,-3+.4,"@"), triangle(-1,-3,-2,4,2,1) 
 )}}}

Now put the point of the compass on the center 
and draw the circle:

{{{drawing(400,400,-10,10,-10,10, locate(-2-.2,4.4,"@"), graph(400,400,-10,10,-10,10), circle(-3/2,1/2,sqrt(25/2)),  locate(-3/2-.2,1/2+.4,"@"),
locate(2-.2,1.4,"@"), locate(-1-.2,-3+.4,"@"), triangle(-1,-3,-2,4,2,1) 
 )}}}

-----------------------

Maybe I'll do the last one tomorrow.  I'm getting sleepy. Check back to
see if I've done it.

Edwin</pre>

c)The sides of a triangle are on the line 6x+7y+11=0,2x-9y+11=0,and 9x+2y-11=0.Find the equation of the cirlce inscribed in the triangle. 
(looking forward for someone who will asnwer this..ty in advance =)