Challenge

[logistic_guy]

Point \(\displaystyle C\) is located at the origin. Line \(\displaystyle \ell\) is tangent to \(\displaystyle \large\odot\)\(\displaystyle C\) at \(\displaystyle (-4, 3)\). Use the diagram to complete the problem.

\(\displaystyle \bold{a.}\) Find the slope of line \(\displaystyle \ell\).
\(\displaystyle \bold{b.}\) Write the equation for \(\displaystyle \ell\).
\(\displaystyle \bold{c.}\) Find the radius of \(\displaystyle \large\odot\)\(\displaystyle C\).
\(\displaystyle \bold{d.}\) Find the distance from \(\displaystyle \ell\) to \(\displaystyle \large\odot\)\(\displaystyle C\) along the \(\displaystyle y\)-axis.

 

[jonah2.0]

Beer induced catatonia follows.

Point \(\displaystyle C\) is located at the origin. Line \(\displaystyle \ell\) is tangent to \(\displaystyle \large\odot\)\(\displaystyle C\) at \(\displaystyle (-4, 3)\). Use the diagram to complete the problem.

\(\displaystyle \bold{a.}\) Find the slope of line \(\displaystyle \ell\).
\(\displaystyle \bold{b.}\) Write the equation for \(\displaystyle \ell\).
\(\displaystyle \bold{c.}\) Find the radius of \(\displaystyle \large\odot\)\(\displaystyle C\).
\(\displaystyle \bold{d.}\) Find the distance from \(\displaystyle \ell\) to \(\displaystyle \large\odot\)\(\displaystyle C\) along the \(\displaystyle y\)-axis.

View attachment 39047

You got this, ...

 

[khansaheb]

Point \(\displaystyle C\) is located at the origin. Line \(\displaystyle \ell\) is tangent to \(\displaystyle \large\odot\)\(\displaystyle C\) at \(\displaystyle (-4, 3)\). Use the diagram to complete the problem.

\(\displaystyle \bold{a.}\) Find the slope of line \(\displaystyle \ell\).
\(\displaystyle \bold{b.}\) Write the equation for \(\displaystyle \ell\).
\(\displaystyle \bold{c.}\) Find the radius of \(\displaystyle \large\odot\)\(\displaystyle C\).
\(\displaystyle \bold{d.}\) Find the distance from \(\displaystyle \ell\) to \(\displaystyle \large\odot\)\(\displaystyle C\) along the \(\displaystyle y\)-axis.

View attachment 39047

show us your effort/s to solve this problem.

 

[The Highlander]

Point \(\displaystyle C\) is located at the origin. Line \(\displaystyle \ell\) is tangent to \(\displaystyle \large\odot\)\(\displaystyle C\) at \(\displaystyle (-4, 3)\). Use the diagram to complete the problem.

\(\displaystyle \bold{a.}\) Find the slope of line \(\displaystyle \ell\).
\(\displaystyle \bold{b.}\) Write the equation for \(\displaystyle \ell\).
\(\displaystyle \bold{c.}\) Find the radius of \(\displaystyle \large\odot\)\(\displaystyle C\).
\(\displaystyle \bold{d.}\) Find the distance from \(\displaystyle \ell\) to \(\displaystyle \large\odot\)\(\displaystyle C\) along the \(\displaystyle y\)-axis.

View attachment 39047

See here.

 

[logistic_guy]

Beer induced catatonia follows.

You got this, ...

Thankslike a piece of Cake?

show us your effort/s to solve this problem.

See here.

Appreciate it


How to attack this Challenge? Start by writing what you have.

Equation of the circle.
\(\displaystyle x^2 + y^2 = r^2\)

Equation of the line.
\(\displaystyle y - 3 = m(x + 4)\)

And thanks to Calculus, we have a third equation for the slope, \(\displaystyle m\).
First consider taking the top part of the circle which has this equation:
\(\displaystyle y = \sqrt{r^2 - x^2}\)

Then,

\(\displaystyle m = \frac{dy}{dx} = \frac{-x}{\sqrt{r^2 - x^2}}\)
Oops we are not allowed to use Calculusso we have to figure out a different method. The problem did not say that, but this problem will be given to you in some point of your life where you probably have not yet learned calculus.

Also it would not deserve to be called a Challenge if you used Calculus.

Let's recap our skills in Geometry. A tangent line to a circle is perpendicular to the radius of the circle. This will help us in this way:

Draw a radius from the center of the circle to the point of tangency. We can find the slope of this line since we have the two points, \(\displaystyle (0,0)\) and \(\displaystyle (-4,3)\).

\(\displaystyle m_r = \frac{3 - 0}{-4 - 0} = -\frac{3}{4}\)

From Algebra, we know that if two lines are perpendicular, the slope of one line is the negative reciprocal of the slope of the other.
So the slope of the line \(\displaystyle \ell\) is:

\(\displaystyle m = -\frac{1}{m_r} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}\)

From this, we can complete the equation of the line \(\displaystyle \ell\).
\(\displaystyle y - 3 = m(x + 4)\)
\(\displaystyle y - 3 = \frac{4}{3}(x + 4)\)
Or
\(\displaystyle y = \frac{4}{3}x + \frac{25}{3}\)

To find the radius of the circle, we can simply use the equation of the circle.
\(\displaystyle x^2 + y^2 = r^2\)
\(\displaystyle (-4)^2 + (3)^2 = r^2\)
\(\displaystyle 16 + 9 = r^2\)
\(\displaystyle r = 5\)

To find the distance from the line \(\displaystyle \ell\) to the circle along the \(\displaystyle y\)-axis, we need first to the find the point where the line \(\displaystyle \ell\) crosses the \(\displaystyle y\)-axis.
\(\displaystyle y = \frac{4}{3}x + \frac{25}{3}\)
Set \(\displaystyle x = 0\).

\(\displaystyle y = \frac{25}{3}\)
We need to find the distance from the top of the circle which is the point \(\displaystyle (0, 5)\) to the point \(\displaystyle \left(0, \frac{25}{3}\right)\).
You can figure out this directly, but let us use the distance formula.
[imath]\displaystyle d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0 - 0)^2 + \left(\frac{25}{3} - 5\right)^2} = \sqrt{\left(\frac{25}{3} - \frac{15}{3}\right)^2}[/imath]

[imath]= \displaystyle \sqrt{\frac{100}{9}} = \frac{10}{3}[/imath]

Bringing them all together:

[imath]\displaystyle \bold{a.} \ m = \frac{4}{3}[/imath]

[imath]\displaystyle \bold{b.} \ y = \frac{4}{3}x + \frac{25}{3}[/imath]

[imath]\displaystyle \bold{c.} \ r = 5[/imath]

[imath]\displaystyle \bold{d.} \ d = \frac{10}{3}[/imath]