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 the line joining the centers.
The two given circles each have radius \(1\) cm and touch each other at \(P\), so their centers and \(P\) all lie on one straight line, with each center exactly \(1\) cm from \(P\) on opposite sides. Since \(PA\) is a diameter of the first circle, \(A\) also lies on this same line, on the far side of the first circle's center from \(P\), with \(PA=2\) cm. Let \(D\) be the center of the second circle; it also lies on this line, \(1\) cm from \(P\) on the opposite side of the first circle. So, along this one line, \(AP=2\) and \(PD=1\), giving \(AD=AP+PD=3\) cm.

Step 2: Place BC.
\(BC\) is a diameter of the second circle, drawn perpendicular to line \(AP\) at the second circle's center \(D\). Since the second circle has radius \(1\) cm, \(BD=DC=1\) cm, with \(BD\perp AD\).

Step 3: Locate the center of the third circle.
The third circle passes through \(A\), \(B\), \(C\). Because \(B\) and \(C\) are placed symmetrically on either side of line \(AD\), being the two ends of a diameter perpendicular to it, the center of the third circle, call it \(G\), must lie on line \(AD\) itself (the perpendicular bisector of chord \(BC\) always passes through a circle's center, and here that bisector is exactly line \(AD\)). Let \(AG=x\), so \(GD=AD-AG=3-x\).

Step 4: Use the fact that G is equidistant from A and B.
Since \(G\) is the center of the circle through \(A\), \(B\), \(C\), all three are at distance \(x\) from \(G\): \(GA=GB=GC=x\). Triangle \(BDG\) has a right angle at \(D\) (since \(BD\perp AD\) and \(G\) lies on \(AD\)), so by the Pythagorean theorem: \[ GB^2 = BD^2+GD^2 \] \[ x^2 = 1^2+(3-x)^2 \]

Step 5: Solve for x. \[ x^2 = 1+9-6x+x^2 \] \[ 0 = 10-6x \] \[ x = \dfrac{10}{6} = \dfrac{5}{3} \]

Final Answer:
The radius of the third circle is \(\dfrac{5}{3}\) cm. \[ \boxed{\dfrac{5}{3}\text{ cm}} \]