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Given that equation of a circle $C$ is $x^2 +y^2 -10x -8y +18=0$

Find the equations of the tangents to the circle $C$ which are parallel to line $y-x+5=0$

(Originally this was not a differentiation question,but I found out that if I restricted myself to pre-calculus,I would have to use discriminant and then the quadratic formula.)

$x^2 +y^2 -10x -8y +18=0$

By differentiating with respect to $x$,

$2x +2y y' -10 -8y'=0$

Hence,

$y'=\frac{-2x+10}{2y-8}$

As the gradient is 1,

$2y-8=-2x+10$

$y=-x+9$

That's where I got stuck.As the gradient is $1$,why does last equation has a gradient of $-1$?Where did I go wrong?Lastly,is there any other easier way ?

Edit: Subst. $y=-x+9$ into $C$

$x^2 + (9-x)^2 -10x-8(9-x)+18=0$

$2x^2 +81-18x -10x -72+8x+18=0$

$2x^2 -20x -27=0$

This would give me $x=\frac {10+\sqrt{154}}{2} $or $x=\frac {10-\sqrt{154}}{2}$

Hence,$y=\frac {8\pm\sqrt{154}}{2}$

So,the equation is $y=x-1\pm\sqrt{154}$

However when I did it the pre-calculus way,the answer is $ y=x-1 \pm \sqrt{46}$

Shub
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Arc Neoepi
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    You didn't go wrong anywhere: that line's equation is telling you on what line the points that fulfill $;y'=1;$ are. You now have to check where that line intersects the circle in order to find out the points and thus the equations... – DonAntonio Sep 22 '16 at 09:50
  • I have added further workings to the questions to the final step,however the answer is different from the pre-calculus way of doing it. – Arc Neoepi Sep 22 '16 at 10:13
  • You have an arithmetic mistake: $;81-72+18=\color{red}+27;$ , and you wrote $;-27;$ . But for that your on the right track – DonAntonio Sep 22 '16 at 10:43
  • Related: http://math.stackexchange.com/questions/1917790/a-line-tangent-to-a-circle/1919090#1919090. In fact there's no need for any differentiation. – Michael Hoppe Sep 22 '16 at 10:46

4 Answers4

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You have to complete your work.

Take $y=-x+9$ in the equation of the circle that becomes: $$ 2x^2-20x+27=0 $$ solve for $x$ and find the two roots $x_1$ and $x_2$. Find $y_1=-x_1+9$ and $y_2=-x_2+9$ . The points $(x_1,y_1)$ and $8x_2,y_2)$ are the points of tangency. Now write the equations of the two lines of slope $m=1$ through these two points.

Note that your solution is wrong because you have a $-27$ in the equation. The correct solution for $x$ is: $$ x=\frac{10\pm \sqrt{46}}{2} $$

The other method, without derivative, is to search a line of equation $y=x+q$ that is tangent to the circle. This means that the system $$ \begin{cases} y=x+q\\ x^2+y^2-10x-8y+18=0 \end{cases} $$ has only one solution. So, substitute $y$ in the second equation so that you have a second degree equation in $x$ with a parameter $q$ and find the value of the parameter $q$ such that the discriminat of this equation is null $\Delta(q)=0$. Since $\Delta(q)$ is a second degree polynomial in $q$ you find two values for $q$ that are the intercepts of the the two tangent lines.

Another method is to intersect the line orthogonal to the $y-x+5=0$ and passing through the center of the circle with the circle. The common points are the points of tangency.

Emilio Novati
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You have used differentiation alright, however by an invalid next step you got the straight line that joins two points of tangentcy.

Apart from what others correctly indicated, another way is to solve for two $C$s from a quadratic equation of intersection with the variable intercept straight line.

$$ y = x + C,$$

setting discriminant to zero to capture tangency.

Edit1:

Your shortcut procedure gives rise to a join of two points of same slope known as a conjugate diameter.

Try to sketch the ellipse and conjugate diameter for example

$$ x^2 + 2 y^2 -10 x -8 y +18 = 0 $$

By the above procedure

$$ \frac{dy}{dx}= \frac{5-x}{2y-4}$$

when the slope is set to $1$ and simplified it gives the conjugate diameter

$$ x + 2y =9 $$

which cuts all segments of lines of same slope$ =1$ included between the ellipse segments ( or circle).

Please check some more cases or google images of conjugate diameter.

Narasimham
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  • Invalid next step?What do you mean? – Arc Neoepi Sep 23 '16 at 09:05
  • By that step you do not get 2 lines parallel to given line, but get $ x+y = 9 $ which is a straight line joining two tangent points. – Narasimham Sep 23 '16 at 09:55
  • Can you explain why I got the $x+y=9$? Instead of 2 straight lines – Arc Neoepi Sep 23 '16 at 09:59
  • With a single slope you get many parallel lines of same slope. To get two straight lines at once you need a single second order quadratic equation factorizable separately into two starlight lines. To get separate lines of same slope you need two different intercepts $C$. Your procedure gives a line which is a join of the tangent points. – Narasimham Sep 23 '16 at 10:49
  • thanks for the insight.Really helped me understand better! – Arc Neoepi Sep 23 '16 at 11:16
  • (contd) .... known as the conjugate diameter. – Narasimham Sep 23 '16 at 12:07
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I think you dont have to use differentiation in this case. You just have to solve for the inverse tangent of the line in order to get the perpendicular line that passes through the given line as well as the center of the circle. After that, find the point of intersection of that circle and the perpendicular line. Use system of equations to solve that. After that you found the point of tangency at the circle, use the slope of the given line then use point slope form in order to get the equation of the line parallel to the given line that is tangent to the circle.

Jhan Prix Naval
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An equation of circle can be written as

$$(x - a)^2 + (y - b)^2 = r^2$$

In our case $a = 5, b = 4, r = \sqrt{23}$

Let the tangent touch the circle at $(x_1, y_1) = (a + r \cosθ, b + r \sinθ)$.

Slope at this point $m = \frac{dy}{dx} = \frac{dy}{dθ} \frac{dθ}{dx} = -\cot θ$

We know this will be equal to slope of given line $m = 1$

So $θ = \frac{-π}4,\frac{3π}4$.

The equation of tangent will be $$\frac{y - y_1}{x - x_1} = 1$$

Plugging in the values of $y_1$ and $x_1$, $$y = x - 1 \pm \sqrt{46}$$

Shub
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