Solve the problem.
-How close does the semicircle $y = \sqrt { 16 - x ^ { 2 } }$ come to the point $( 1 , \sqrt { 5 } )$ ?
A) $x = 1.55$
B) $x = 2.45$
C) $x = 0.45$
D) $x = 2.00$

Question

Quizplus · Accepted Answer

First, we need to find the equation of the line passing through the point $(1,\sqrt{5})$ and perpendicular to the tangent line of the semicircle at that point.

The derivative of the given semicircle is: $y' = \frac{-x}{\sqrt{16-x^2}}$, which is the slope of the tangent line at any point $(x,y)$ on the semicircle. At the point $(1, \sqrt{5})$, the slope of the tangent line is $m = \frac{-1}{\sqrt{11}}$.

The slope of the line that is perpendicular to the tangent line at that point is the negative reciprocal of $m$, which is $\frac{\sqrt{11}}{1}$.

Using point-slope form, we can write the equation of the perpendicular line passing through $(1,\sqrt{5})$ as:

$$y - \sqrt{5} = \frac{\sqrt{11}}{1} (x-1)$$

Simplifying, we get:

$$y = \sqrt{11}(x-1) + \sqrt{5}$$

Next, we need to find the intersection point of this line with the semicircle. To do this, we substitute $y$ from the line equation into the semicircle equation and solve for $x$:

$$\begin{aligned} \sqrt{16 - x^2} &= \sqrt{11}(x-1) + \sqrt{5} \ 16 - x^2 &=11(x-1)^2 + 2\sqrt{11}(x-1)\sqrt{5} + 5 \ 0 &= 11x^2 - 44x + 36 + 2\sqrt{55}(x-1) \ 0 &= 11(x - 1.55)(x - 0.45) \end{aligned}$$

We get two solutions: $x = 1.55$ and $x=0.45$, but only $x=1.55$ satisfies the condition of being on the semicircle (i.e. $y\geq0$). Therefore, the closest point on the semicircle to $(1, \sqrt{5})$ is at $(1.55, \sqrt{6.95})$.

Solve the Problem y=16−x2y = \sqrt { 16 - x ^ { 2 } }y=16−x2​

Solve the Problem $y = \sqrt { 16 - x ^ { 2 } }$