Problem 72 - Algebra and Trigonometry with Analytic Geometry

Question

A circle \(C_{1}\) of radius 5 has its center at the origin. Outside this circle is a first-quadrant circle \(C_{2}\) of radius 2 that is tangent to \(C_{1}\). The \(y\)-coordinate of the center of \(C_{2}\) is 3 . Find the \(x\)-coordinate of the center of \(C_{2}\).

Step-by-Step Solution

Answer

The \(x\)-coordinate of the center of \(C_{2}\) is \(2\sqrt{10}\).

Step 1: Understanding Tangency Conditions

Circle \(C_{1}\) with center at the origin \((0, 0)\) and radius 5 is tangent to circle \(C_{2}\) with radius 2. If two circles are tangent externally, the distance between their centers is equal to the sum of their radii. For this problem, this distance will be \(5 + 2 = 7\).

Step 2: Set Up the Distance Equation

Let the center of circle \(C_{2}\) be \((x, 3)\), since the \(y\)-coordinate is given as 3. Using the distance formula, the distance between centers \((0, 0)\) and \((x, 3)\) is \(\sqrt{x^2 + 3^2}\). Set this distance equal to 7: \(\sqrt{x^2 + 9} = 7\).

Step 3: Solve for \(x\)

Square both sides of the equation from Step 2: \(x^2 + 9 = 49\). Subtract 9 from both sides to get: \(x^2 = 40\). Take the square root of both sides: \(x = \sqrt{40}\). Simplify: \(x = 2\sqrt{10}\).

Step 4: Check Quadrant Consistency

Since circle \(C_{2}\) is in the first quadrant, which means both \(x\) and \(y\) are positive, the \(x\)-coordinate \(x = 2\sqrt{10}\) must be positive, and it is. Thus, the solution is consistent with the stated conditions.

Key Concepts

Understanding Tangent Circles

Tangent circles might sound complicated, but they are just two circles that touch at exactly one point. This concept is important because it tells us how to determine the distance between their centers. In this case, we have Circle \( C_1 \) centered at the origin \((0,0)\) with a radius of 5. Circle \( C_2 \) is outside Circle \( C_1 \), and has a smaller radius of 2. Because they are tangent, they just gently touch without overlapping or gapping, ensuring the distance between their centers is exactly the sum of their radii.

So, the distance from center to center is calculated as the radius of Circle \( C_1 \) plus the radius of Circle \( C_2 \):

• Radius of \( C_1 \) is 5
• Radius of \( C_2 \) is 2

Adding these together gives us a distance of 7.

Utilizing the Distance Formula

The distance formula is a handy tool for finding the straight-line distance between two points in a coordinate plane. It comes from the Pythagorean theorem and is written as \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) for two points \((x_1, y_1)\) and \((x_2, y_2)\). In our problem, we need to find the distance between the centers of Circle \( C_1 \) at \((0, 0)\) and Circle \( C_2 \) at \((x, 3)\).

Plug each coordinate into the formula:

• First point \((x_1, y_1)\): \((0, 0)\)
• Second point \((x_2, y_2)\): \((x, 3)\)

The distance becomes \(\sqrt{x^2 + 3^2}\) since \((0 - x)^2 = x^2\) and \((3 - 0)^2 = 9\). Since these circles are tangent, we've already determined this distance should be 7. So we equate the formula to 7 and solve for \(x\), giving us \(\sqrt{x^2 + 9} = 7\).

Position in the First Quadrant

The location of a point in the coordinate plane's first quadrant means both its \(x\) and \(y\) coordinates are positive. For Circle \( C_2 \), the \(y\)-coordinate of its center is given as 3, and since we're told it's in the first quadrant, we must find a positive \(x\)-coordinate. This consistency check is crucial to make sure solutions don't stray into negative territories for this quadrant.

After setting up and solving the distance equation, we find \(x^2 = 40\), leading to \(x = \sqrt{40}\) or simply \(x = 2\sqrt{10}\). Since \(2\sqrt{10} \approx 6.32\) is positive, it fits perfectly within the constraints of being in the first quadrant.

Having the \(x\)-coordinate positive indicates that Circle \( C_2 \) does sit properly in the first quadrant, confirming the circle's correct placement and ensuring that all conditions of tangency and positioning are satisfied.