Problem 74 - Algebra 2

Question

What is the distance between \(T(9,-5)\) and the center of the circle with equation \((x-6)^{2}+(y+1)^{2}=10 ?\)

Step-by-Step Solution

Answer

5 units

Step 1: Identify the center of the circle

The general form of a circle's equation is \((x-a)^2 + (y-b)^2 = r^2\), where \((a,b)\) is the center of the circle. In the exercise, the equation of the circle is \((x-6)^2+(y+1)^2=10\). Comparing it to the general form, we can see that the center of the circle is \((a,b) = (6,-1)\).

Step 2: Use the distance formula

The formula to calculate the distance between two points \((x1,y1)\) and \((x2,y2)\) is \(\sqrt{(x2-x1)^2 + (y2-y1)^2}\). The first point is \(T(9,-5)\) and the second point is the center of the circle, which is \((6,-1)\). Plugging these in the formula, we get \(\sqrt{(6-9)^2 + (-1-(-5))^2}\). This simplifies to \(\sqrt{(-3)^2 + 4^2}\) and then to \(\sqrt{9 + 16}\).

Step 3: Calculate the distance

Now, sum up the numbers under the square root to find the distance, \(\sqrt{25} = 5\). Thus, the distance between point \(T(9,-5)\) and the center of the circle \((6,-1)\) is 5 units.

Key Concepts

Equation of a Circle

The equation of a circle is an essential concept in geometry, because it helps us understand the shape's unique characteristics. A circle can be defined in a plane by its center and radius. Generally, the equation of a circle is written as \[(x-a)^2 + (y-b)^2 = r^2\]where

• \((a, b)\) represents the center of the circle, and
• \(r\) is the radius of the circle.

By breaking down this formula, you can find information unique to each circle, like its specific center and radius length. This helps in any problem-solving situations involving circles, like finding distances from points or areas.

Center of a Circle

The center of a circle is the point from which every point on the circle's perimeter is equidistant. Understanding how to find the center of a circle from its equation is crucial. When given an equation \[(x-a)^2 + (y-b)^2 = r^2\]comparing it to the equation, we know that the coordinate \((a, b)\) is the center. So if you see an equation like \((x-6)^2+(y+1)^2=10\):

• The center would be \((6, -1)\).

This information reveals not just location, but aids in further computations, like finding distances or plotting the circle.

Solving Algebraic Equations

Solving algebraic equations involves finding the values of variables that make the equation true. When it comes to circles, we often deal with equations that include squares and square roots. For instance, finding the distance between a point \((x_1, y_1)\) and the center of a circle \((x_2, y_2)\) can be achieved using the distance formula \[\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]. By isolating the variables and solving for unknown values, algebraic equations empower us to navigate real-world problems involving geometric shapes effectively.