Question

The circle given by *2 + y2 - 4x -10=0 can be written in standard form 2. like this: (x - h) + y2 = 14 What is the value of h in this equation?

Answer 1

Given the equation of the circle:

`x^2+y^2-4x-10=0`

We will complete the square for the terms of x

so,

`\begin{gathered} (x^2-4x)+y^2=10 \\ (x^2-4x+4)+y^2=10+4 \\ (x-2)^2+y^2=14 \end{gathered}`

compare the final result with the equation:

`(x-h)^2+y^2=14`

So, the value of h = 2

So, the answer will be:

`h=2`