Hi. I'm struggling to find the coordinates of C. I've been getting the equations of the line with A as y=-x/2 - 5. And the equation of the line with B y = -x + 2/10. with A(-2,-4) and B(/10,/10). I get the x coordinate of C as 2(5+ 2/10), but obviously, this is incorrect as it is a positive. Can someone explain where I'm going wrong!
Tangent to a circle
- Thread starter freyatown
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Hi. I'm struggling to find the coordinates of C. I've been getting the equations of the line with A as y=-x/2 - 5. And the equation of the line with B y = -x + 2/10. with A(-2,-4) and B(/10,/10). I get the x coordinate of C as 2(5+ 2/10), but obviously, this is incorrect as it is a positive. Can someone explain where I'm going wrong!
Dr.Peterson
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Are you using "/10" to mean "√10"? Don't do that; it's very confusing. If you can't paste in the right symbols (or learn Latex), use "sqrt(10)".
The picture is not only "not drawn accurately"; it is positively misleading (no pun intended). The slopes of the lines are such that the intersection C is in the fourth quadrant, not the second!
The picture is not only "not drawn accurately"; it is positively misleading (no pun intended). The slopes of the lines are such that the intersection C is in the fourth quadrant, not the second!
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Consider the circle centered at the origin given by:
[MATH]x^2+y^2=r^2[/MATH]
Now, also consider also a point on the circle \(\left(x_1,y_1\right)\). To find the line tangent to the circle, let us define the line as:
[MATH]y=m(x-x_1)+y_1[/MATH]
We know the slope of this line must be:
[MATH]m=-\frac{x_1}{y_1}[/MATH]
And so the tangent line is:
[MATH]y=-\frac{x_1}{y_1}(x-x_1)+y_1[/MATH]
We are told that [MATH]r^2=20[/MATH], and that the \(y\)-coordinate of point A is -4:
[MATH]x_1=\pm\sqrt{20-(-4)^2}=\pm2[/MATH]
If point A is in quadrant 3, then [MATH]x_1=-2[/MATH], and our tangent line is:
[MATH]y=-\frac{-2}{-4}(x-(-2))+(-4)=-\frac{1}{2}(x+2)-4=-\frac{1}{2}x-5[/MATH]
This agrees with what you found. Likewise for the other tangent line, I find:
[MATH]y_1=\pm\sqrt{20-(\sqrt{10})^2}=\pm\sqrt{10}[/MATH]
If point B is in quadrant 1, then [MATH]y_1=\sqrt{10}[/MATH], and our tangent line is:
[MATH]y=-\frac{\sqrt{10}}{\sqrt{10}}(x-\sqrt{10})+\sqrt{10}=-(x-\sqrt{10})+\sqrt{10}=-x+2\sqrt{10}[/MATH]
Based on Dr. Peterson's reply, I assume this is what you meant. Graphing the circle and the two lines, we see:
As Dr. Peterson pointed out, point C will be in the 4th quadrant.
[MATH]x^2+y^2=r^2[/MATH]
Now, also consider also a point on the circle \(\left(x_1,y_1\right)\). To find the line tangent to the circle, let us define the line as:
[MATH]y=m(x-x_1)+y_1[/MATH]
We know the slope of this line must be:
[MATH]m=-\frac{x_1}{y_1}[/MATH]
And so the tangent line is:
[MATH]y=-\frac{x_1}{y_1}(x-x_1)+y_1[/MATH]
We are told that [MATH]r^2=20[/MATH], and that the \(y\)-coordinate of point A is -4:
[MATH]x_1=\pm\sqrt{20-(-4)^2}=\pm2[/MATH]
If point A is in quadrant 3, then [MATH]x_1=-2[/MATH], and our tangent line is:
[MATH]y=-\frac{-2}{-4}(x-(-2))+(-4)=-\frac{1}{2}(x+2)-4=-\frac{1}{2}x-5[/MATH]
This agrees with what you found. Likewise for the other tangent line, I find:
[MATH]y_1=\pm\sqrt{20-(\sqrt{10})^2}=\pm\sqrt{10}[/MATH]
If point B is in quadrant 1, then [MATH]y_1=\sqrt{10}[/MATH], and our tangent line is:
[MATH]y=-\frac{\sqrt{10}}{\sqrt{10}}(x-\sqrt{10})+\sqrt{10}=-(x-\sqrt{10})+\sqrt{10}=-x+2\sqrt{10}[/MATH]
Based on Dr. Peterson's reply, I assume this is what you meant. Graphing the circle and the two lines, we see:
As Dr. Peterson pointed out, point C will be in the 4th quadrant.
Dr.Peterson
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Did you notice that there are two possible points for each of A and B? Since we know they got the picture wrong with regard to the quadrant of C, we can't trust it for the choice of A and B.
I graphed all the possibilities and found that their solution is for A(+2. -4), and B(√10, +√10). There are three other solutions, one of which is the one we all got based on your assumptions.
This may be one of the worst problems I've seen in terms of ambiguity and misleading pictures.
I graphed all the possibilities and found that their solution is for A(+2. -4), and B(√10, +√10). There are three other solutions, one of which is the one we all got based on your assumptions.
This may be one of the worst problems I've seen in terms of ambiguity and misleading pictures.
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