Question

In Problems 17-22, find the center and radius of the circle with the given equation. 4 x^{2}+16 x+15+4 y^{2}+6 y=0

Short answer

Center: (-2, -3/4), Radius: \(\frac{\sqrt{5}}{4}\)

Step-by-step solution

Group Terms

The given equation is \(4x^2 + 16x + 15 + 4y^2 + 6y = 0\). First, we group the terms involving \(x\) and \(y\) separately: \( (4x^2 + 16x) + (4y^2 + 6y) + 15 = 0\).

Factor Out Coefficients of Quadratic Terms

Factor out the coefficient of the quadratic terms in each group. For \(x\) terms, factor out 4: \(4(x^2 + 4x)\). For \(y\) terms, factor out 4: \(4(y^2 + \frac{3}{2}y)\).

Complete the Square for x

To complete the square for \(x^2 + 4x\), add and subtract \((\frac{4}{2})^2 = 4\) inside the parentheses: \(4((x^2 + 4x + 4) - 4) = 4((x + 2)^2 - 4)\).

Complete the Square for y

To complete the square for \(y^2 + \frac{3}{2}y\), add and subtract \((\frac{3}{4})^2 = \frac{9}{16}\) inside the parentheses: \(4((y^2 + \frac{3}{2}y + \frac{9}{16}) - \frac{9}{16}) = 4((y + \frac{3}{4})^2 - \frac{9}{16})\).

Simplify Equation

Substitute complete squares and simplify:\[4((x + 2)^2 - 4) + 4((y + \frac{3}{4})^2 - \frac{9}{16}) + 15 = 0\]. Distribute the 4 and combine all constants:\[4(x + 2)^2 - 16 + 4(y + \frac{3}{4})^2 - \frac{9}{4} + 15 = 0\].

solve for final equation

Combine all like terms:\[4(x + 2)^2 + 4(y + \frac{3}{4})^2 - 16 - \frac{9}{4} + 15 = 0\] becomes:\[4(x + 2)^2 + 4(y + \frac{3}{4})^2 = 1.25\]. Then divide by 4:\[(x + 2)^2 + (y + \frac{3}{4})^2 = \frac{1.25}{4}\].

find center and radius

The equation now has the form \((x - h)^2 + (y - k)^2 = r^2\), where the center \((h, k)\) is \((-2, -\frac{3}{4})\) and radius \(r\) is \(\sqrt{\frac{1.25}{4}} = \frac{\sqrt{1.25}}{2}\). Simplify to find the radius: \(\frac{\sqrt{5}}{4}\)

Key concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial, making equations easier to work with, especially in a circle's equation.
It involves adding and subtracting a certain number in order to form a complete square.
This technique is essential for graphing and converting equations into the circle's standard form:

• For any quadratic term like \(x^2 + bx\), add and subtract \((\frac{b}{2})^2\) inside the equation.
• Factor any leading coefficient first to make the process simpler.
• Use this transformation to convert quadratic expressions into forms like \((x-h)^2\).

In the provided exercise, after factoring out constants, we completed the square separately for both the \(x\) and \(y\) terms.
This restructuring helps in identifying the center and radius of the circle by making it easier to compare with the equation's standard form.

Center of a Circle

The center of a circle in its equation's standard form \[(x-h)^2 + (y-k)^2 = r^2\] is identified as the point \((h, k)\).Finding the center is pivotal for understanding the circle's position in a coordinate plane.
Here's how we determine it:

• After completing the square for both \(x\) and \(y\) sets, the transformation will position them as \((x-h)^2\) and \((y-k)^2\).
• The center coordinates \((h, k)\) are simply the values opposite the constants added inside the completed squares.
  For instance, in the equation resulting from our problem, \((x+2)^2 + (y+\frac{3}{4})^2\), the center becomes \((-2, -\frac{3}{4})\).

Understanding the center’s coordinates helps in many graphing applications, offering a precise look at where the circle is exactly located on the plane.

Radius of a Circle

The radius of a circle is a fundamental part of its geometry.
In the equation \[(x-h)^2 + (y-k)^2 = r^2\], \(r^2\) represents the radius squared.
The calculation of the radius involves:

• Converting the constant on the right side of the equation by factoring out and simplifying as necessary.
• Taking the square root of the resulting number to determine \(r\).
• For the example equation, after simplifying, we reached the form \( (x + 2)^2 + (y + \frac{3}{4})^2 = \frac{5}{16} \), with \(r^2 = \frac{5}{16}\).
• To find \(r\), compute the square root of \( \frac{5}{16} \), which results in \( \frac{\sqrt{5}}{4} \).

Knowing the radius not only helps in plotting the circle accurately but also in calculating key properties like circumference and area.
Remember, the radius extends from the center to any point on the circle, representing the circle's constant distance from the center.