- Thread starter
- Admin
- #1
Welcome to our community
Be a part of something great, join today!
ivyrianne's question at Yahoo! Answers regarding finding a circle
- Thread starter MarkFL
- Start date
- Thread starter
- Admin
- #2
Hello again ivyrianne,
Let's let the equation of the circle be:
\(\displaystyle (x-h)^2+(y-k)^2=r^2\)
We have 3 parameters to determine: $h,k,r$.
In order for the center of the circle to lie on the line $4x+3y=2$, we require:
(1) \(\displaystyle 4h+3k=2\)
Now, let's rewrite the tangent lines in slope-intercept form:
a) \(\displaystyle y=-x-4\)
b) \(\displaystyle y=7x+4\)
If we have a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$, then the perpendicular distance $d$ from the point to the line is given by:
\(\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}\)
You may find derivations of this formula here:
http://www.mathhelpboards.com/f49/finding-distance-between-point-line-2952/
Using this formula for $d=r$ and the two tangent lines, we find that we require:
\(\displaystyle r=\frac{|-h-4-k|}{\sqrt{2}}=\frac{|7h+4-k|}{5\sqrt{2}}\)
Using the definition \(\displaystyle |x|\equiv\sqrt{x^2}\), we may square through and write:
\(\displaystyle \left(5(h+k+4) \right)^2=(7h-k+4)^2\)
Solving (1) for $k$, we find:
\(\displaystyle k=\frac{2-4h}{3}\)
Substituting, there results:
\(\displaystyle \left(5\left(h+\frac{2-4h}{3}+4 \right) \right)^2=\left(7h-\frac{2-4h}{3}+4 \right)^2\)
Combining like terms, we find:
\(\displaystyle \left(\frac{5}{3}(h-14) \right)^2=\left(\frac{5}{3}(5h+2) \right)^2\)
Hence, we must have:
\(\displaystyle (h-14)^2-(5h+2)^2=0\)
\(\displaystyle (6h-12)(-4h-16)=0\)
\(\displaystyle -24(h-2)(h+4)=0\)
\(\displaystyle h=-4,\,2\)
Case 1: \(\displaystyle h=-4\)
\(\displaystyle k=\frac{2-4(-4)}{3}=6\)
\(\displaystyle r=\frac{\left|-(-4)-4-6 \right|}{\sqrt{2}}=3\sqrt{2}\)
Case 2: \(\displaystyle h=2\)
\(\displaystyle k=\frac{2-4(2)}{3}=-2\)
\(\displaystyle r=\frac{\left|-2-4-(-2) \right|}{\sqrt{2}}=2\sqrt{2}\)
Thus, the two possible circles are:
\(\displaystyle \left(x+4 \right)^2+\left(y-6 \right)^2=18\)
\(\displaystyle \left(x-2 \right)^2+\left(y+2 \right)^2=8\)
Here is a plot of the three lines and the two circles:
Let's let the equation of the circle be:
\(\displaystyle (x-h)^2+(y-k)^2=r^2\)
We have 3 parameters to determine: $h,k,r$.
In order for the center of the circle to lie on the line $4x+3y=2$, we require:
(1) \(\displaystyle 4h+3k=2\)
Now, let's rewrite the tangent lines in slope-intercept form:
a) \(\displaystyle y=-x-4\)
b) \(\displaystyle y=7x+4\)
If we have a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$, then the perpendicular distance $d$ from the point to the line is given by:
\(\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}\)
You may find derivations of this formula here:
http://www.mathhelpboards.com/f49/finding-distance-between-point-line-2952/
Using this formula for $d=r$ and the two tangent lines, we find that we require:
\(\displaystyle r=\frac{|-h-4-k|}{\sqrt{2}}=\frac{|7h+4-k|}{5\sqrt{2}}\)
Using the definition \(\displaystyle |x|\equiv\sqrt{x^2}\), we may square through and write:
\(\displaystyle \left(5(h+k+4) \right)^2=(7h-k+4)^2\)
Solving (1) for $k$, we find:
\(\displaystyle k=\frac{2-4h}{3}\)
Substituting, there results:
\(\displaystyle \left(5\left(h+\frac{2-4h}{3}+4 \right) \right)^2=\left(7h-\frac{2-4h}{3}+4 \right)^2\)
Combining like terms, we find:
\(\displaystyle \left(\frac{5}{3}(h-14) \right)^2=\left(\frac{5}{3}(5h+2) \right)^2\)
Hence, we must have:
\(\displaystyle (h-14)^2-(5h+2)^2=0\)
\(\displaystyle (6h-12)(-4h-16)=0\)
\(\displaystyle -24(h-2)(h+4)=0\)
\(\displaystyle h=-4,\,2\)
Case 1: \(\displaystyle h=-4\)
\(\displaystyle k=\frac{2-4(-4)}{3}=6\)
\(\displaystyle r=\frac{\left|-(-4)-4-6 \right|}{\sqrt{2}}=3\sqrt{2}\)
Case 2: \(\displaystyle h=2\)
\(\displaystyle k=\frac{2-4(2)}{3}=-2\)
\(\displaystyle r=\frac{\left|-2-4-(-2) \right|}{\sqrt{2}}=2\sqrt{2}\)
Thus, the two possible circles are:
\(\displaystyle \left(x+4 \right)^2+\left(y-6 \right)^2=18\)
\(\displaystyle \left(x-2 \right)^2+\left(y+2 \right)^2=8\)
Here is a plot of the three lines and the two circles: