Question

Two circles of radius \(1\) cm each touch each other at point \(P\). A third circle is drawn through the points \(A\), \(B\) and \(C\) such that \(PA\) is a diameter of the first circle, and \(BC\) (perpendicular to \(AP\)) is a diameter of the second circle. The radius of the third circle is:

• A. 9/5 cm
• B. 7/4 cm
• C. 5/3 cm
• D. 10/2 cm

Hint:

Drop a perpendicular from the center of the third circle onto line AP and apply the Pythagorean theorem in the small right triangle formed.

Answer 1

The Correct Option is C

Step 1: Set up coordinates along the line of centers.
Place the tangency point at $P=(0,0)$, with line $AP$ along the x-axis. Since $PA$ is a diameter of the first circle (radius $1$), put $A=(-2,0)$, so the first circle's center sits at the midpoint $(-1,0)$, exactly $1$ unit from $P$.

Step 2: Place the second circle's center.
The second circle (radius $1$) also passes through $P$ and touches the first circle there, so its center $D$ sits $1$ unit from $P$ on the opposite side, giving $D=(1,0)$.

Step 3: Place B and C.
$BC$ is a diameter of the second circle, perpendicular to the x-axis, through $D=(1,0)$. Since the radius is $1$, the endpoints are $B=(1,1)$ and $C=(1,-1)$.

Step 4: Find the circumcenter of triangle ABC directly.
Because $B$ and $C$ are mirror images across the x-axis, and $A$ sits on the x-axis, the whole configuration is symmetric about the x-axis, so the center of the circle through $A$, $B$, $C$ must also lie on the x-axis; call it $G=(g,0)$.

Step 5: Set the distance from G to A equal to the distance from G to B. $$GA = |g+2|, \quad GB = \sqrt{(g-1)^2+1}$$ Setting $GA=GB$ and squaring: $$(g+2)^2 = (g-1)^2+1$$ $$g^2+4g+4 = g^2-2g+2$$ $$6g = -2$$ $$g = -\frac{1}{3}$$

Step 6: Compute the radius. $$GA = \left|-\frac{1}{3}+2\right| = \frac{5}{3}$$

Final Answer:
The radius of the third circle is $\frac{5}{3}$ cm, matching the value found by the synthetic Pythagorean method through a completely independent coordinate calculation. $$\boxed{\frac{5}{3}\text{ cm}}$$