Circles
Tangent circles and chord geometry
MJAT_TS1_P2
Grade 12
Question:
Let $O$ be the centre of a circle and $AE$ be a chord. The diameter $AP$ is drawn, and $M$ is the foot of the perpendicular from $E$ to $AP$. A smaller circle is drawn touching $EM$, $AP$ at $D$, and internally tangent to the circle with diameter $AP$. Then the value of $\dfrac{AD}{AE}$ equals
Step-by-Step Solution
Key Concept: Let the circle with diameter $AP$ have centre $O'$ and radius $R$. The angle in a semicircle gives $\angle AEP = 90°$. Since $EM \perp AP$, triangles $AEM$ and $APE$ are similar: $AE^2 = AM \cdot AP$.
From similar triangles $\triangle AEM \sim \triangle APE$: $AE^2 = AM \cdot AP$. The small circle of radius $r$ touches $AP$ at $D$, so $AM = AD - r$ (since $M$ is where $EM$ meets $AP$, and the circle touches $EM$). Also $(R-r)^2 = r^2 + (AD-R)^2$, giving $AE = AD$. So $\dfrac{AD}{AE} = \mathbf{1}$.
Correct Answer: 1