Circles \(C_1\), \(C_2\), and \(C_3\), with centers \(O_1\), \(O_2\), and \(O_3\), and radii \(r_1\), \(r_2\), and \(r_3\), respectively, touch each other as shown in the following figure. Given \(r_1 = 2\) cm, \(r_2 = 1\) cm and the angle \(\angle O_1O_3O_2\) is \(90^{\circ}\), \(r_3 =\) _____ cm.
Show Hint
The right angle at \(O_3\) makes \(O_1O_2\) the hypotenuse of triangle \(O_1O_3O_2\); apply the Pythagorean theorem with \(O_1O_2=3\), \(O_1O_3=2+r_3\), \(O_2O_3=1+r_3\).
Step 1: Translate "touch each other" into distances between centers.
When two circles touch each other externally (each circle outside the other, just touching at one point), the distance between their centers equals the sum of their radii. Applying this to each pair here:
\[ O_1O_2 = r_1 + r_2 \]
\[ O_1O_3 = r_1 + r_3 \]
\[ O_2O_3 = r_2 + r_3 \]
Step 2: Substitute the known radii.
We are given \(r_1 = 2\) cm and \(r_2 = 1\) cm, so:
\[ O_1O_2 = 2 + 1 = 3 \]
\[ O_1O_3 = 2 + r_3 \]
\[ O_2O_3 = 1 + r_3 \]
Step 3: Use the given right angle in triangle \(O_1O_3O_2\).
We are told the angle \(\angle O_1O_3O_2\), the angle of the triangle at vertex \(O_3\), is \(90^{\circ}\). In any triangle, the side opposite the right angle is the longest side and is called the hypotenuse. Since the right angle sits at \(O_3\), the side opposite it is \(O_1O_2\), so \(O_1O_2\) is the hypotenuse of this right triangle, and the Pythagorean theorem applies with the other two sides, \(O_1O_3\) and \(O_2O_3\), as the legs:
\[ O_1O_2^2 = O_1O_3^2 + O_2O_3^2 \]
Step 5: Solve the quadratic in \(r_3\).
\[ 2r_3^2 + 6r_3 + 5 - 9 = 0 \]
\[ 2r_3^2 + 6r_3 - 4 = 0 \]
Dividing every term by 2:
\[ r_3^2 + 3r_3 - 2 = 0 \]
Apply the quadratic formula, \(r_3 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\), with \(a=1\), \(b=3\), \(c=-2\):
\[ r_3 = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2} \]
Step 6: Pick the physically valid root.
A radius cannot be negative, so reject the root that gives a negative value. Since \(\sqrt{17} \approx 4.12\), the root \(\dfrac{-3-\sqrt{17}}{2}\) is negative and must be discarded. The remaining root is:
\[ r_3 = \frac{-3+\sqrt{17}}{2} = \frac{1}{2}\left(-3+\sqrt{17}\right) \approx 0.56 \text{ cm} \]
This is a small positive radius, which makes sense: \(C_3\) is the little circle tucked into the corner between the two bigger circles \(C_1\) and \(C_2\).
Step 7: Why the other options are wrong.
Option (B), \(\frac{1}{2}(3+\sqrt{17})\), uses the wrong sign in front of the 3 and gives a value bigger than \(r_2\), too large for a circle squeezed into that corner. Option (C), \(\frac{1}{2}(-2+\sqrt{17})\), comes from mistakenly using 2 instead of 3 as the constant term when rearranging the quadratic. Option (D), \(\frac{1}{2}(-3+2\sqrt{17})\), doubles the square-root term by mistake, for example by forgetting to divide the whole quadratic by 2 before applying the formula.
Final Answer:
\[ r_3 = \frac{1}{2}\left(-3+\sqrt{17}\right) \text{ cm} \]
Find dV/dt