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Find the equation of the circle tangent to the $x$-axis at the origin and tangent to the line $4x-3y+24=0$.

My Approach:

Let the equation of the required circle be: $$x^2+y^2+2gx+2fy+c=0$$

Let the equation to the tangent at origin $(0,0)$ to the above circle be $gx+fy+c=0$.

Then, what should I do? please help me to continue.

Thanks in Advance.

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  • $\begingroup$ The wording is confusing (at least to me); Does "touching the $\;x\,-$ axis at the origin" = the circle's tangential to the $\;x\,-$ axis at the origin? $\endgroup$ – DonAntonio Oct 8 '16 at 11:39
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A circle tangent to $x$-axis in $(0,0)$ has its center in $C_0(0,y_0)$ and its radius is $|y_0|$. Thus its equation has the form:

$$\tag{1}x^2+(y-y_0)^2=y_0^2$$

The distance from the center $C_0(x_0,y_0)=(0,y_0)$ of the circle to the straight line is (see (http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html)):

$$\frac{|4x_0-3y_0+24|}{\sqrt{4^2+(-3)^2}}=\frac{|-3y_0+24|}{5}$$

This distance must be equal to the radius $|y_0|$ giving the following equation:

$$\frac{|-3y_0+24|}{5}=|y_0| \ \ \Leftrightarrow \ \ -3y_0+24=\pm 5 y_0$$

which has 2 solutions: $y_0=3$ or $y_0=-12$ giving, (see (1)):

  • either the equation $x^2+(y-3)^2=3^2$ i.e., $x^2+y^2-6y=0$ (center $(0,3)$, radius 3),

  • or the equation $x^2+(y+12)^2=12^2$ i.e., $x^2+y^2+24y=0$ (center $(0,-12)$, radius = 12).

enter image description here

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  • $\begingroup$ Hmm, how can this approach be modified to give the second solution? $\endgroup$ – Ian Miller Oct 8 '16 at 11:32
  • $\begingroup$ @JeanMarie, The question has two equations: $x^2+y^2+24y=0$ and $x^2+y^2-6y=0$. What do you have to say about that? $\endgroup$ – pi-π Oct 8 '16 at 11:34
  • $\begingroup$ You are right. I correct it $\endgroup$ – Jean Marie Oct 8 '16 at 11:34
  • $\begingroup$ Where did you get the two equations? $\endgroup$ – pi-π Oct 8 '16 at 11:38
  • $\begingroup$ Ah! The inclusion of the absolute value gives the second solution. Nicely done @JeanMarie. $\endgroup$ – Ian Miller Oct 8 '16 at 11:41
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As the circle touches the x-axis at (0,0) then the centre of the circle must be on the y-axis and the y coordinate of the centre must be equal to the radius. So the centre is $(0,r)$. So the equation is:

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

Then you want to find the point of intersection of the circle and the straight line. Lets rearrange the straight line:

$$x=\frac{3y-24}{4}$$

Sub in:

$$\left(\frac{3y-24}{4}\right)^2+(y-r)^2=r^2$$

Multiple by $4^2$ to get rid of the fractions:

$$(3y-24)^2+16(y-r)^2=16r^2$$

Expand:

$$9y^2-144y+576+16y^2-32yr+16r^2=16r^2$$

Collect and sort by powers of $y$:

$$25y^2-16(2r+9)y +576=0$$

Now we expect there to be only one point of intersection so this quadratic should have only one solution. This occurs when the discriminant is zero.

$$(16(2r+9))^2-4\cdot25\cdot576=0$$

Pull out the $16^2$ term and factorise $576$ in preparation to divide out common factor.

$$256(2r+9)^2-4\cdot25\cdot64\cdot9=0$$

Divide by common factor of $256$.

$$(2r+9)^2-25\cdot9=0$$

Expand and multiply:

$$4r^2+36r+81-225=0$$

Collect:

$$4r^2+36r-144=0$$

Divide by 4:

$$r^2+9r-36=0$$

Factorise:

$$(r-3)(r+12)=0$$

Solve:

So $r=3$ or $r=-12$

Note that $r=-12$ is a valid solution. It means that the circle is below the axis where earlier we assumed it was above the axis (when we set the centre to $(0,r)$).

So the two solutions are: $x^2+(y-3)^2=3^2$ and $x^2+(y+12)^2=12^2$

Or in expanded form: $x^2+y^2-6y=0$ and $x^2+y^2+24y=0$

enter image description here

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    $\begingroup$ I like this, its first principles and gets all the answers. $\endgroup$ – user24142 Oct 8 '16 at 11:36
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    $\begingroup$ There are many ways to approach this problem. I set out a technique which I thought would be suitable for your current level of knowledge based on the small start you made. $\endgroup$ – Ian Miller Oct 8 '16 at 11:38
  • $\begingroup$ What is meant by this: " Now we expect there to be only one point of intersection so this quadratic should have only one solution. This occurs when the discriminant is zero."? $\endgroup$ – pi-π Oct 8 '16 at 16:07
  • $\begingroup$ As the line is a tangent there is only one point of intersection with the circle. Secondly if a quadratic only has one solution then the discriminant is equal to zero $\Delta=b^2-4ac=0$ $\endgroup$ – Ian Miller Oct 8 '16 at 17:19

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