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I would love some help on finding the equation of the circles tangent to $d_1, d_2$ and $d_3$, given $$\begin{cases}d_1: y=4x-10 \\ d_2: y=9/4x-15/4 \\ d_3: y=3x-15 \end{cases} $$

My approach: I know that $d_2$ and $d_3$ are parallel. The circles, in order to be tangent to the $3$ lines must pass by $M$( intersection between $d_1$ and $d_2$) and $N$ (intersection between $d_1$ and $d_3$).

I found the coordinates of these points. Then, I found the radius, as it is the distance between $ d_2$ and $d_3$. But in the end, it appears that I am wrong, but I dont understand why...

Many thanks in advance

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  • $\begingroup$ I don't think d2 and d3 are parallel. Hint: the center of the circle is equidistant from the 3 tangents. $\endgroup$ – Gribouillis Oct 16 '17 at 17:13
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    $\begingroup$ No pair of lines are parallel. You need to use the distance of a point to a line formula to find the equation of the circle. Additionally, we have multiple solutions to the given problem. $\endgroup$ – Math Lover Oct 16 '17 at 17:16
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Let $(a,b)$ be a center of the circle.

Thus, $$\frac{|4a-b-10|}{\sqrt{17}}=\frac{|9a-4b-60|}{\sqrt{97}}$$ and $$\frac{|4a-b-10|}{\sqrt{17}}=\frac{|3a-b-15|}{\sqrt{10}}.$$ Now solve this system an you'll get four centers.

For all center find the distance between the center and some our line.

You'll get a radius of the circle and write equations of these circles.

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Given any line, $y=mx+b$, then a $\bot$ is given by $y=(1/m)(h-x)+k$ where (h,k) is a point not on the line which the perpendicular will pass thru. In our case, (h,k) will be the center point of the circle(s). You can derive the above equation for yourself by simply using the perpendicular slope of $-1/m$ and substituting in h and k to a line equation.

Since $(h,k)$ was specified to not be on the line, then we can solve the two equations simultaneously and find a point that is on the line. That intersection is given by $$x_1=\frac{m(k-b)+h}{m^2+1}$$ $$y_1=\frac{b+hm+km^2}{m^2+1}$$ Now we have two points, $(h,k)$, the circle center, and $(x_1,y_1)$ a point on one of the 3 lines, whichever one you got $m$ and $b$ from. The distance between these 2 points is given by $$\frac{|h\cdot m+b-k|}{\sqrt{m^2+1}}$$ which also turns out to be a general formula for the distance from any point $(h,k)$ to any line $y=mx+b$.

Using specific lines as given by the problem we can equate the distances, which are the radii of each circle (however many there are). $$\frac{|4h - k - 10|}{ \sqrt{4^2+1}} = \frac{|3h -k - 15| }{\sqrt{3^2+1}}$$ $$\frac{|4h-10-k|}{\sqrt{4^2+1}} = \frac{|(9/4)h-15/4-k|}{\sqrt{(9/4)^2+1}}$$

These two equations were a little more complicated to solve than expected. Wolfram alpha gave up since I don't have the paid version, but a little persistence came up with these centers and radii. $$(696.07,-234.15)\;\;729.65$$ $$(-7.939,-29.144)\;\;3.05$$ $$(12.729,30.881\;\;2.43$$ $$(4.855,3.8436)\;\;1.35$$

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The distance to a straight line of equation $$ax+by+c=0$$ is obtained as

$$\frac{|ax+by+c|}{\sqrt{a^2+b^2}}.$$

The center of a circle tangent to the three lines is equidistant to them and you need to solve

$$\dfrac{|ax+by+c|}{\sqrt{a^2+b^2}}= \dfrac{|a'x+b'y+c'|}{\sqrt{a'^2+b'^2}}= \dfrac{|a''x+b''y+c''|}{\sqrt{a''^2+b''^2}}.$$

Removing the absolute values, you have actually four systems of two equations in two unknows, giving four distinct solutions (they are the inscribed circle and three escribed ones).

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