Finding radius of a circle from a tangent and a secant

Question

We know how long the tangent line is and I found out that DC is 4, which would make EC 10. I can't figure out what to do next. I've tried to use similar triangles (BEA and BAC) and found no success.

Answer 1

By the power of a point: $$ AB^2=BC\cdot BE\implies BE=\frac{AB^2}{BC}=18\implies EC=BE-BC=10. $$ Recalling that the diameter drawn through the midpoint of a chord is perpendicular to the chord one obtains by Pythagorean theorem for the circle radius $R$ the value: $$ R^2=PD^2-\left(DE-\frac12EC\right)^2+\left(\frac12EC\right)^2=4^2-(6-5)^2+5^2=40.\tag1$$

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For explanation of (1) see figure below. I hope the relations: $$EQ=QC=\frac12EC;\quad QD=ED-EQ;\\ PQ^2=PD^2-QD^2;\quad PC^2=PQ^2+QC^2$$ are clear enough.

Answer 2

According to the chord theorems for the intersection points both inside and outside the circle, we have

$$DE\cdot CD = (r-PD) \cdot ( r + PD)\implies 6CD=(r-4)(r+4) $$

$$BA^2 = BC \cdot ( BC + CD + DE)\implies 12^2=8(8+CD+6)$$

Eliminate $CD$ to obtain $r = 2\sqrt{10}$.

Answer 3

$\triangle PDC $ and $\triangle EPC $ are similar. Because both tiangles are iscoseles . So( $\frac{4}{r}=\frac{r}{10}$ ) $r= 2\sqrt10$