Find the equation of the circle

[anil kumar]

A circle touches the line 2x + 3y +1 = 0 at the pt. (1,-1) and is orthogonal
to the circle whose one pair of diametrically opposite end pts. are (3,0) and
(1,-3).Find the equation of the circle.

 

[AKG]

You know the diameter of the second circle is $\sqrt{(3-1)^2 + [0 - (-3)]^2} = \sqrt{13}$. Therefore, the radius, $r = \sqrt{13}/2$, and $r^2 = 13/4$

Now the second circle can be represented by the equation:

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

$$(x - a)^2 + (y - b)^2 = 13/4$$

You can make two equations with two unknowns, and solve for a and b:

$$(1):\ \ (3-a)^2 + b^2 = 13/4$$

$$(2):\ \ (1-a)^2 + (3+b)^2 = 13/4$$

$$a = 2,\ b = -1.5$$

This should help you get started.

 

[AKG]

Now, let (m,n) be the center of the circle we are solving for. We know:

$$(1-m)^2 + (1+n)^2 = R^2$$

where R is the radius of the circle we're solving for. We also know that there must be points on the other circle, $(x_1,y_1)$ and [itex](x_2,y_2)[/tex] such that:

$$(x_1 - 2,y_1 + 1.5) \cdot (x_1 - m, y_1 - n) = 0$$

$$(x_2 - 2,y_2 + 1.5) \cdot (x_2 - m, y_2 - n) = 0$$

We also have:

$$(x_1 - 2)^2 + (y_1 + 1.5)^2 = 13/4$$

$$(x_2 - 2)^2 + (y_2 + 1.5)^2 = 13/4$$

and:

$$\frac{1-m}{-1-n} = 3/2$$

and:

$$(x_1-m)^2 + (y_1-n)^2 = R^2$$

$$(x_2-m)^2 + (y_2-n)^2 = R^2$$

8 equations, 7 unknowns. Looks ugly, but possible.

 

[loseyourname]

Okay, perhaps someone can explain to me why this approach didn't work.

I started out with the three pieces of information that we have to begin with:

1) The equation of the circle is $$(x - a)^2 + (y - b)^2 = 13/4$$

2) The equation of the tangent line is $$2x + 3x + 1 = 0$$

3) The line is tangent to the circle at the point $$(1,-1)$$

In other words, $$\frac{dy}{dx}$$ at $$(1,-1)$$ is $$-2/3$$

So I differentiated the equation of the circle to obtain the equation for $$\frac{dy}{dx}$$

$$\frac{d}{dx}(x - a)^2 + \frac{d}{dx}(y - b)^2 = \frac{d}{dx}13/4$$

If $$u = x - a$$ and $$v = y - b$$,

then $$u\prime = 1$$ and $$v\prime = 1$$

I rewrote the equation as

$$\frac{d}{dx}u^2 + \frac{d}{dx}v^2 = \frac{d}{dx}13/4$$

$$2uu\prime + 2vv\prime = 0$$

$$2(x - a) + 2(y - b)\frac{dy}{dx} = 0$$

Solving for $$\frac{dy}{dx}$$, I got

$$2(y - b)\frac{dy}{dx} = -2(x - a)$$

$$\frac{dy}{dx} = \frac{-2(x - a)}{2(y - b)} = -\frac{(x - a)}{(y - b)}$$

I then substituted in the known values for $$(x,y)$$ and $$\frac{dy}{dx}$$,

$$(1,-1)$$ and $$-2/3$$

$$-2/3 = -\frac{(1 - a)}{(1 - b)}$$

The negatives cancel and we get

$$1 - a = 2$$ and $$-1 - b = 3$$

$$a = -1$$ and $$b = -4$$

Substituting these values back into the equation of the circle, we get

$$(x + 1)^2 + (y + 4)^2 = 13/4$$

Now this equation is not correct, as it never touches the point $(1,-1)$. I'm sure I'm overlooking something very simple, but it's 1:30 AM here and I'm really tired. Can someone help me out?

 

[Muzza]

Yes, you are overlooking something very simple: the circle we're asked to find doesn't (necessarily) have a radius of sqrt(13/4).

The circle with "diametrically opposite end pts. (3,0) and (1,-3)" has a radius of sqrt(13/4). Its equation is easy to find, since you know two points on it. AKG did it in his first post. This circle is not a tangent to 2x + 3y + 1 = 0.

What we're looking for is a totally different circle (that IS a tangent to 2x + 3y + 1 = 0). The only connection this new circle has with the old circle, is that they are orthogonal to each other.

*edit* That's not to say that you can't find a circle with a radius of sqrt(13/4) that is tangent to 2x + 3y + 1 = 0 at (1, -1). Let the circle's equation be (x - a)^2 + (y - b)^2 = 13/4. Work out its intersection with the tangent line. The discriminant must be zero. You now have an equation relating a and b. But since you also know that (1, -1) must lie on the circle, you have another equation with a and b. Two equations, two variables, should be solvable. In fact, if you slog through the calculations (or enlist a computer to do it for you ;)), you'll find that the equation is x^2 + (y + 5/2)^2 = 13/4 or (x - 2)^2 + (y - 1/2)^2 = 13/4.

You can apply the same technique (letting the circle's equation be (x - a)^2 + (y - b)^2 = r^2, though) to solve the original problem, only you have to apply the condition given on Mathworld... Very un-pretty though.

And now I see the mistake you did. If you have this equation:

$$-2/3 = -\frac{(1 - a)}{(1 - b)}$$

1 - a is not necessarily equal to 2, and 1 - b is not necessarily equal to 3. Since $$\frac{2}{3} = \frac{4}{6}$$, is 2 = 4 and 3 = 6?

 

[HallsofIvy]

A lot of the calculations you did are not necessary: The circle such that (3, 0) and (1, -3) has radius $\frac{\sqrt{13}}{2}$ AND center at ((3+1)/2,(0-3)/2)=
(2, -3/2) so THAT circle is (x-2)²+ (y+ 3/2)²= 13/4.

The circle you want to find must be of the form (x-a)²+ (y-b)²= R² (any circle is of that form). If it "touches" (is tangent to) the line
2x+ 3y+ 1= 0 then it must be tangent at (x,y) such that 2x+ 3y+ 1= 0 and
x²+ y²= R². That's two equations for 3 unknowns.

In order to be perpendicular to the circle (x-2)²+ (y+ 3/2)²= 13/4, The must be two points (x,y) satisfying the equations of both circles and such that the derivatives y'= (y- 3/2)/(x- 2)= (y-b)/x-a). You need to solve all those equations for a and b.