areacirclestangent line
Above is the picture in question. A circle is given, center (-2,4) and a point outside the circle (0,10) is shown. Asked to calculate the area of the quadrilateral ABCD, I figured that this kite has sides 2 (the radii of the circle) and 6 (the difference between (0,4) and (0,10)).
How can I calculate the area of the kite?
(I tried calculating the diagonals and got rad40 for one of the diagonals, but cannot figure out how to calculate the other)
Thanks!
A useful technique for this situation which avoids using calculus goes like this:
Find the equation of a line through the point $(4,-3)$ with gradient $k$, and then use this (linear) equation to substitute into the equation of the circle to get a quadratic. The two roots of this quadratic would give the 2 points of intersection of the line with the circle, but in our case, we want to know the values of $k$ for which the quadratic has a double root (i.e. the 2 points of intersection are actually coincident). Use the standard condition ($b^2=4ac$) for a double root of a quadratic.
In your case, the equation of the line through $(4,-3)$ with gradient $k$ is $y+3 = k(x-4)$, or $y+7 = k(x-4)+4$. Substituting this into the equation of the circle gives:
$$(x+2)^2 + (k(x-4)+4))^2 = 25$$
So you need to rearrange this quadratic in $x$ and find out which vales of $k$ give double roots.
Let $R$ and $r$ be the radii of the bigger and smaller circles respectively. The required area is $\pi r^2$.
To follow your approach using "the tangents meeting are equal theory", consider the common tangent line of the two circles:
$$r + R = \sqrt{R^2+R^2} = R\sqrt 2$$
Alternatively, I use the distance from the outside right-angle to the tangent point to calculate both radii:
$$\begin{align*} R + \sqrt{R^2+R^2} = R \left(1+\sqrt2\right) &= BM\\ r + \sqrt{r^2+r^2} = r \left(1+\sqrt2\right) &= R \end{align*}$$
Best Answer
Note that, $AB=AD=6$ units (tangents from a point to a circle are equal in length), and $BC=DC=2$(radius) units. Join $AC$. $\triangle ACD$ is congruent to triangle $\triangle ABC$ (tangents make $90^0$ with the radius, AB=AD, BC=DC, hence they're congruent by the RHS criteria). Hence,
total area $= 2*1/2*CD*AD=2*6=12$ unit square.