Chapter 1.4, Problem 42AYU

To determine

To find: The standard form of the equation of the circle

Answer to Problem 42AYU

The standard form of the equation of the circle is ${\left(x-4\right)}^2+{\left(y+2\right)}^2=25$

Explanation of Solution

Given:

Center $(4,-2)$ and tangent to the line $x=1$

Calculation:

A tangent to a circle is a line that touches the circle at a particular point $\left(x,y\right)$

The center of the given circle $\left(h,k\right)$ is $(4,-2)$

The line $x=1$ is a vertical line with the $x-intercept$ as $1$ and no $y-intercept$

Since the circle is a tangent to the line, it touches the line at the point $(-1,2)$ .

Therefore, the coordinates $\left(x,y\right)$ can be written as $(-1,2)$

If $\left(x,y\right)$ represents the coordinates of any point on the circle with radius $r$ and center $\left(h,k\right)$ , then by the distance formula, the radius of the circle is equivalent to the distance between the points $\left(x,y\right)$ and $\left(h,k\right)$

  $\sqrt{{\left(x-h\right)}^2-{\left(y-k\right)}^2}=r$

Substitute $x=1,y=-2,h=4,k=-2$

  $r=\sqrt{{\left(1+4\right)}^2+{\left(-2-(-2)\right)}^2}$

  $=\sqrt{25}$

  $=5$

Therefore, the radius of the circle is $5$

The standard form of the circle is given by ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

  ${\left(x-4\right)}^2+{\left(y-(-2)\right)}^2=5^2$

  ${\left(x-4\right)}^2+{\left(y+2\right)}^2=25$

Conclusion:

Hence, the standard form of the equation of the circle is ${\left(x-4\right)}^2+{\left(y+2\right)}^2=25$