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Prove It · #2 July 3rd, 2009, 04:53 AM

Quote:

Originally Posted by helloying

find the eqn of the circle that passes through the point A(8,1) and B(7,1) and has , for its tangent at B, the line 3x-4y-21=0

The equation of a circle is given by

$(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the centre and $r$ is the radius.

From the information given, we can generate three equations in three unknowns, so that we can solve them simultaneously for $h, k$ and $r$ .

Substituting point A $(8, 1)$ into the equation gives

$(8 - h)^2 + (1 - k)^2 = r^2$ .

Substituting point B $(7, 1)$ into the equation gives

$(7 - h) + (1 - k)^2 = r^2$ .

This circle's derivative is given by:

$\frac{d}{dx}[(x - h)^2 + (y - k)^2] = \frac{d}{dx}(r^2)$

$\frac{d}{dx}[(x - h)^2] + \frac{d}{dx}[(y - k)^2] = 0$

$2(x - h) + \frac{dy}{dx}\cdot\frac{d}{dy}[(y - k)^2] = 0$

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

$2(y - k)\frac{dy}{dx} = -2(x - h)$

$\frac{dy}{dx} = -\frac{x - h}{y - k}$ .

This is the derivative, which gives us the gradient of the tangent at all points on the circle.

If it's tangent at B is the line $3x - 4y - 21 = 0$ , we can rearrange this to read $y = \frac{3}{4}x - \frac{21}{4}$ .

So the gradient of the tangent at point B is $\frac{3}{4}$ .

So at point $(7, 1)$ the gradient is $\frac{3}{4}$ .

Substituting these values into the derivative gives

$-\frac{7 - h}{1 - k} = \frac{3}{4}$

$\frac{h - 7}{1 - k} = \frac{3}{4}$

$h - 7 = \frac{3}{4}(1 - k)$

$h - 7 = \frac{3}{4} - \frac{3}{4}k$

$h = \frac{31}{4} - \frac{3}{4}k$ .

We can now solve for $h, k$ and $r$ .

So far we have:

$(8 - h)^2 + (1 - k)^2 = r^2$ and $(7 - h)^2 + (1 - k)^2 = r^2$ .

Therefore

$(8 - h)^2 + (1 - k)^2 = (7 - h)^2 + (1 - k)^2$

$(8 - h)^2 = (7 - h)^2$

$64 - 16h + h^2 = 49 - 14h + h^2$

$15 = 2h$

$h = \frac{15}{2}$ .

We also know:

$h = \frac{31}{4} - \frac{3}{4}k$ , so

$\frac{15}{2} = \frac{31}{4} - \frac{3}{4}k$

$\frac{3}{4}k = \frac{1}{4}$

$k = \frac{1}{3}$ .

Finally, we know that

$(8 - h)^2 + (1 - k)^2 = r^2$

$\left(8 - \frac{15}{2}\right)^2 + \left(1 - \frac{1}{3}\right)^2 = r^2$

$\left(\frac{1}{2}\right)^2 + \left(\frac{2}{3}\right)^2 = r^2$

$\frac{1}{4} + \frac{4}{9} = r^2$

$r^2 = \frac{25}{36}$

$r = \frac{5}{6}$ .

Now we finally have enough information for the equation of the circle:

$\left(x - \frac{15}{2}\right)^2 + \left(y - \frac{1}{3}\right)^2 = \left(\frac{5}{6}\right)^2$ .

The following users thank Prove It for this useful post:
helloying
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