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Circle with two tangent lines

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!

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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.

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  • $\begingroup$ This is very clear. Would there be a way of calculating the length of the other diagonal? $\endgroup$ – bagrut1 Jun 6 at 14:47
  • $\begingroup$ @DaniellaLejtman You can mark it correct if it answers your question. And yes, that can also be done. Consider $\triangle BCD$ and $\triangle ABD$ and find BD. It's a bit tedious. AC is easy to find out since $\triangle ACD$ is a right-angled triangle. Now, $Area= AC.BD/2$ $\endgroup$ – Tapi Jun 6 at 14:59
  • $\begingroup$ It'd actually be easier to calculate BD once you find out the area. Use the formula $A=(1/4)b\sqrt{4a^2−b^2}$ for $\triangle BCD$ and $\triangle ABD$. @DaniellaLejtman They have the same value of $b$ (which we're finding out) but different $a$ values, i.e. $2$ and $6$. Now add the areas of those two triangles and equate it with the area of $ABCD$ (which is $12$). It'd take a lot of time if you have to calculate the diagonal $BD$. $\endgroup$ – Tapi Jun 7 at 14:33
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Since $\overline{AB}$ is tangent to the circle, $m\angle B = 90°$, so the quadrilateral consists of two right-triangles. Both triangles share a common hypotenuse $\overline{AC}$ and have a congruent leg, as both are the radius of the circle. Hence, the two right-triangles are congruent, so $A_{kite} = 2A_{\triangle} = 2\cdot\frac{1}{2}bh = bh$. From the diagram, $b = 0-(-2) = 2$ and $h = 10-4 = 6$, so $A_{kite} = 2\cdot 6 = 12$ (in square units).

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Since $AB=AD$ and $BC=CD$, and the radii are perpendicular to the tangent lines, the triangles $ABC$ and $ACD$ are equal. The area of $ACD$ is $(AC\cdot CD)/2$. Then the area of quadrilateral is $AC\cdot CD$.

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  • $\begingroup$ Should be AD⋅𝐶𝐷 $\endgroup$ – Ido Sarig Jun 15 at 18:55

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