Question

the circle given by ( x^{2}+y^{2}-4x - 10 = 0 ) can be written in standard form like this: ( (x - h)^{2}+y^{2}=14 ). what is the value of ( h ) in this equation? ( h=)______

Explanation:

Step1: Complete the square for the x - terms

Given \(x^{2}+y^{2}-4x - 10=0\). Group the x - terms: \((x^{2}-4x)+y^{2}=10\).
Using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), for \(x^{2}-4x\), if \(a = x\) and \(2ab=4x\) (so \(b = 2\)), then \(x^{2}-4x=(x - 2)^{2}-4\).
The equation becomes \((x - 2)^{2}-4+y^{2}=10\).

Step2: Simplify the equation

Add 4 to both sides of the equation \((x - 2)^{2}-4+y^{2}=10\).
We get \((x - 2)^{2}+y^{2}=14\).

Answer:

\(2\)