Problem 47 - Algebra 2

Question

A circle has its center on the line with equation \(y=2 x .\) It passes through \((1,-3)\) and has a radius of \(\sqrt{5}\) units. Write an equation of the circle.

Step-by-Step Solution

Answer

The equation of the circle is \((x + 1)^2 + (y + 2)^2 = 5\).

Step 1: Understand the Circle Equation

The general equation of a circle with center \((h, k)\) and radius \(r\) is: \[(x - h)^2 + (y - k)^2 = r^2\]We need to find \((h, k)\) and know that the radius \(r = \sqrt{5}\).

Step 2: Set Up Center Coordinates

Since the circle's center is on the line \(y = 2x\), the center can be represented as \((h, 2h)\) for some value of \(h\).

Step 3: Use Point on the Circle

The point \((1, -3)\) lies on the circle. Using the circle's equation, substitute \((x, y) = (1, -3)\) and \((h, k) = (h, 2h)\):\[(1 - h)^2 + (-3 - 2h)^2 = (\sqrt{5})^2\]This simplifies to:\[(1 - h)^2 + (-3 - 2h)^2 = 5\]

Step 4: Simplify the Equation

Expand and simplify:\[(1 - h)^2 = 1 - 2h + h^2\] \[(-3 - 2h)^2 = 9 + 12h + 4h^2\]Substitute back into the circle equation:\[1 - 2h + h^2 + 9 + 12h + 4h^2 = 5\]Simplify:\[5h^2 + 10h + 10 = 5\]\[5h^2 + 10h + 5 = 0\]

Step 5: Solve Quadratic Equation for h

Factor out the quadratic equation:\[h^2 + 2h + 1 = 0\]This can be factored as:\[(h + 1)^2 = 0\]Thus, \(h = -1\).

Step 6: Identify the Center of the Circle

Substitute \(h = -1\) into the center coordinates \((h, 2h)\):\((h, k) = (-1, -2)\).

Step 7: Write the Equation of the Circle

Substitute \((h, k) = (-1, -2)\) and \(r = \sqrt{5}\) back into the circle equation:\[(x + 1)^2 + (y + 2)^2 = 5\]This is the equation of the circle.

Key Concepts

Circle Center

The center of a circle is one of the most crucial aspects when dealing with the equation of a circle. The general form of a circle’s equation is \[(x - h)^2 + (y - k)^2 = r^2\],where

• \((h, k)\) is the center of the circle, and
• \(r\) is the radius.

In this specific problem, the circle's center must lie on the line defined by the equation \(y = 2x\). This condition lets us express the center in terms of a single variable, \[\text{center} = (h, 2h)\].Since any point on this line satisfies the equation, substituting it into the circle equation along with the known point \((1, -3)\),we can solve for \(h\) to find the exact center of this specific circle.

Radius of a Circle

The radius is always positive and defines the size of the circle. It measures the distance from the center to any point on the edge of the circle. The problem specifies the radius as \(\sqrt{5}\).Plugging this value into the general circle equation \[(x - h)^2 + (y - k)^2 = r^2\],where \(r = \sqrt{5}\),leads us to write \[(x - h)^2 + (y - k)^2 = 5\].Once we determine the circle center from the given conditions, the radius allows us to utilize the equation to include specific points on the circle, like \((1, -3)\),ensure they satisfy this equation. Hence, it is research into the center’s placement and checks with given points.

Quadratic Equation

Quadratic equations come into play when we need to solve for unknowns, such as the circle's center. In this context, after inserting the point and the circle's condition, we obtain the equation\[(1 - h)^2 + (-3 - 2h)^2 = 5\].Expanding and simplifying, we derive a quadratic expression:\[5h^2 + 10h + 5 = 0\].Solving this equation involves either factoring or using the quadratic formula, \[-b \pm \sqrt{b^2 - 4ac} \over 2a\].Here, since \[(h + 1)^2 = 0\],we find \(h = -1\).This result provides the \(x\)-coordinate of the circle center, crucial for constructing the circle's final equation.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves analyzing geometrical figures using a coordinate system. For circles, this means representing them through an equation relating \(x\) and \(y\). Understanding the intersection of algebraic and geometric concepts enables us to determine circle properties, such as the center and the radius from given conditions or geometrical relations, like \[y = 2x\].This line indicates how the center relates within the plane. Using these coordinates and algebraic manipulation, we solve for specific components like \((h, k)\).Coordinate geometry translates visual spatial problems into solvable mathematics using precise equations, as demonstrated in constructing the circle equation \[(x + 1)^2 + (y + 2)^2 = 5\].This approach anchors abstract geometrical relations into comprehensible algebraic solutions.