Problem 7 - Contemporary Precalculus

Question

Find the center and radius of the circle whose equation is given. $$x^{2}+y^{2}+8 x-6 y-15=0$$

Step-by-Step Solution

Answer

Question: Determine the center and radius of the circle given by the equation \(x^2 + 8x + y^2 - 6y = -15\). Answer: The center of the circle is (-4, 3) and the radius is √10.

Step 1: Rewrite the equation

First, let's group the x terms and y terms together: $$(x^2+8x)+(y^2-6y)=-15$$

Step 2: Complete the square for x terms

To complete the square for the \(x\) terms, we need to find the square of half of the linear coefficient of \(x\). Half of 8 is 4, and its square is 16. Add 16 to both sides: $$(x^2+8x+16)+(y^2-6y)=-15+16 \quad \Rightarrow \quad (x+4)^2-16+(y^2-6y)=1$$

Step 3: Complete the square for y terms

Similarly, for the \(y\) terms, half of -6 is -3, and its square is 9. Add 9 to both sides: $$(x+4)^2-16+(y^2-6y+9)=1+9 \quad \Rightarrow \quad (x+4)^2+(y-3)^2=10$$

Step 4: Identify the center and radius

Now, the equation is in the standard form of a circle's equation \((x-a)^2+(y-b)^2=r^2\). By comparing the coefficients, we can identify center \((a,b)=(-4,3)\) and radius \(r=\sqrt{10}\). So, the center of the circle is (-4, 3) and the radius is \(\sqrt{10}\).