Question 610913
{{{x^2+(y-5)^2=25}}}
First let's simplify/ Squaring y-5:
{{{x^2+y^2-10y+25=25}}}
Subtracting 25 from each side we get:
{{{x^2+y^2-10y=0}}}<br>
Now let's convert to polar form. Replacing x with {{{r*cos(theta)}}} and y with {{{r*sin(theta)}}} this becomes:
{{{(r*cos(theta))^2 + (r*sin(theta))^2 - 10(r*sin(theta)) = 0}}}
Simplifying:
{{{r^2*cos^2(theta) + r^2*sin^2(theta) - 10r*sin(theta) = 0}}}
Factoring out {{{r^2}}} from the first two terms:
{{{r^2(cos^2(theta) + sin^2(theta)) - 10r*sin(theta) = 0}}}
Since {{{cos^2(theta) + sin^2(theta) = 1}}}:
{{{r^2(1) - 10r*sin(theta) = 0}}}
which simplifies to:
{{{r^2 - 10r*sin(theta) = 0}}}
Since r cannot be zero, we can divide both sides by r giving:
{{{r - 10sin(theta) = 0}}}
This may be an acceptable answer. Or you could add {{{10sin(theta)}}} to each side:
{{{r = 10sin(theta)}}}