Question

The equation of the circle passing
through the point
(2 + 5 cos thita, -3+ 5 sin thita)
is​

Answer 1

Step-by-step explanation:

circle has center at the Cartesian point (12,5) and radius 13.

Give the Cartesian and polar forms of its equation.

The Cartesian form is (x - h)2 + (y - k)2 = a2, where (h,k) is the center and a is the radius.

Here we have h = 12, k = 5 and a = 13, so we get:

(x - 12)2 + (y - 5)2 = 132,

x2 - 24x + 144 + y2 - 10y + 25 = 169,

x2 + y2 - 24x - 10y = 169 - 144 - 25 = 0,

x2 + y2 - 24x - 10y = 0.

Note that the circle passes through the origin, since if x = 0 and y = 0, the equation is satisfied.

Translating to polars we have $x = r\cos(\theta)$, $y = r\sin(\theta)$, $x^2 + y^2 = r^2(\cos^2(\theta) + \sin^2(\theta) = r^2$, giving:

\begin{displaymath}r^2 - 24r\cos(\theta) - 10r\sin(\theta) = 0, \end{displaymath}

\begin{displaymath}r = 0, \hspace{10pt} r = 24\cos(\theta) + 5\sin(\theta).\end{displaymath}

But the second equation includes the point with r = 0, when $\tan(\theta) = -\frac{24}{5}$, so the first equation is redundant and we may write the polar equation as:

\begin{displaymath}r = 24\cos(\theta) + 5\sin(\theta).\end{displaymath}

Sketch the circle.

Find the equation of the tangent line to the circle at the origin and sketch this line.

The center is at (12,5), so the radius line joining the center to the origin has slope $\frac{5 - 0}{12 - 0} = \frac{5}{12}$.

Then the tangent at the origin is perpendicular to the radius line, so has slope $\frac{-5}{12}$.

So by the point slope form of the equation of a line, the required tangent line has the equation: $y - 0 = -\frac{5}{12}(x - 0)$, or 12y + 5x = 0.