+99 how close does the circle with the radius 4 and center at (4,3) come to the point (10,3)?
+2446 Problem #1: "how close does the line 2x-5y=4 come to the point (1,5)"
When measuring distance from a point to a line, it is important that you find the perpendicular distance. I think a diagram could be useful for this problem. I have provided you with one that encompasses this exact problem:
\(\overline{BC}\) is included in the equation given, \(2x-5y=4\)
d is the perpendicular distance from the desired point, A, to the line,
I think that converting the equation into slope-intercept (y=mx+b) form is the way to go. To do this, solve for y:
| \(2x-5y=4\) | Subtract the x-term to the left-hand side of the equation. |
| \(-5y=-2x+4\) | Divide by -5. |
| \(y=\frac{2}{5}x-\frac{4}{5};\\ m_{\overline{CB}}=\frac{2}{5},b=-\frac{4}{5}\) | |
When the equation is in this form, it gives us much more valuable information that standard form does. For example, we now know what the slope is. This can also allow us to find the slope of the perpendicular line because perpendicular lines are always opposite reciprocals of the original slope.
| \(m_{\overline{AB}}=\frac{2}{5}\Rightarrow -\frac{2}{5}\Rightarrow -\frac{5}{2}\) | Now that we know the slope of this line, let's substitute this into the new equation for the segment. |
| \(y=-\frac{5}{2}x+b\) | There are a few methods to determine the "b." I will just substitute the coordinate that I know that lies on this line, (1,5), and solve for b. |
| \(x=1,y=5;\\ 5=-\frac{5}{2}*1+b\) | Solve for b. |
| \(5=-\frac{5}{2}+b\) | Add 5/2 to solve completely. |
| \(b=5+\frac{5}{2}=\frac{10}{2}+\frac{5}{2}=\frac{15}{2}\) | Therefore, the final equation of the line is the following. |
| \(y=-\frac{5}{2}x+\frac{15}{2}\) | |
Why would we want to find the equation of the line AB. Well, I now have a system of equations, so I can figure out where both lines intersect. As a refresher, here are the equation of both lines again. Both lines are already solved for y, so we can utilize the substitution method to figure out the intersection point.:
\(y=\frac{2}{5}x-\frac{4}{5}\\ y=-\frac{5}{2}x+\frac{15}{2}\)
| \(\frac{2}{5}x-\frac{4}{5}=-\frac{5}{2}x+\frac{15}{2}\) | When there are so many fractions in the same problem, I would suggest multiplying by the LCM of the fractions. In this case, that would be 10. |
| \(10(\frac{2}{5}x-\frac{4}{5})=10(-\frac{5}{2}x+\frac{15}{2})\) | Now, simplify! In the meantime, kiss those fractions goodbye! |
| \(2*2x-2*4=5*-5x+5*15\) | This is the beauty of multiplying the entire equation by the LCM. |
| \(4x-8=-25x+75\) | Now, solve for x. |
| \(29x=83\) | This system of equations is ugly, unfortunately. No nice numbers. |
| \(x=\frac{83}{29}\) | |
Unfortuntately, these are not the nicest numbers. We now have to compute the y-coordinate. Substitute in the x-coordinate for one of the two equations.
| \(y=\frac{2}{5}x-\frac{4}{5}\) | x=83/29. |
| \(y=\frac{2}{5}*\frac{83}{29}-\frac{4}{5}\) | It is possible that the best way to compute this by hand is to factor out 2/5 temporarily. |
| \(y=\frac{2}{5}\left(\frac{83}{29}-2\right)\) | Now, do the subtraction outside of the parentheses. |
| \(\frac{83}{29}-2=\frac{83}{29}-\frac{58}{29}=\frac{25}{29}\) | Plug this into the original computation. |
| \(y=\frac{2}{5}*\frac{25}{29}=\frac{2}{1}*\frac{5}{29}=\frac{10}{29}\) | With this, I am able to do some simplification before multipyling thankfully. |
| \(\left(\frac{83}{29},\frac{10}{29}\right)\) | This is the coordinate of the intersection, or point B. |
To find the distance, just use the distance formula! This finds the distance from point A to B, which is also the closest.
| \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) | The two coordinates we are plugging in is \(\left(\frac{83}{29},\frac{10}{29}\right)\) and (1,5) |
| \(d=\sqrt{\left(\frac{83}{29}-1\right)^2+\left(\frac{10}{29}-5\right)^2}\) | I would just plug this into a calculator and solve. |
| \(d=\sqrt{\frac{54^2}{29^2}+\frac{135^2}{29^2}}\) | Combine the fractions together. |
| \(d=\sqrt{\frac{21141}{29^2}}\) | Distribute the square root symbol to the numerator and denominator. |
| \(d=\frac{\sqrt{21141}}{\sqrt{29^2}}\) | This is really convenient because the square root of something squared is itself. We must factor the numerator, though, to ensure it has no common factors. |
| \(d=\frac{\sqrt{3^6*29}}{29}\) | That's a lot of factors of three! Let's bring that to the outside. |
| \(d=\frac{3^3\sqrt{29}}{29}=\frac{27\sqrt{29}}{29}\approx 5.01\text{units}\) | The radical form is the exact answer, and 5.01 units gives you an idea of the approximate distance. |
Problem #2: "how close does the circle with the radius 4 and center at (4,3) come to the point (10,3)?"
I think this question is easier to answer. The center of the circle lies on (4,3). The radius is 4 units long, so it could extend to the point (8,3), which is 2 units away from the desired point (10,3). I know this because both points lie on the same y-coordinate.
