# eqn of circle and tangent

• Jul 3rd 2009, 02:29 AM
helloying
eqn of circle and tangent
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
• Jul 3rd 2009, 02:53 AM
Prove It
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

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

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

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

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

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

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

This circle's derivative is given by:

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

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

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

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

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

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

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 $\displaystyle 3x - 4y - 21 = 0$, we can rearrange this to read $\displaystyle y = \frac{3}{4}x - \frac{21}{4}$.

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

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

Substituting these values into the derivative gives

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

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

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

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

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

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

So far we have:

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

Therefore

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

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

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

$\displaystyle 15 = 2h$

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

We also know:

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

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

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

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

Finally, we know that

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

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

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

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

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

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

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

$\displaystyle \left(x - \frac{15}{2}\right)^2 + \left(y - \frac{1}{3}\right)^2 = \left(\frac{5}{6}\right)^2$.
• Jul 3rd 2009, 04:55 AM
Tesla23
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

centre of circle must lie on perp bisector of AB which is line x = 7.5, so let centre be C be the point (7.5,y)

so we need to find y such that |AC| = distance of C to line 3x-4y-21=0 so:

$\displaystyle (7.5 - 7)^2 + (y - 1)^2 = \frac{(3*7.5 - 4y - 21)^2}{3^2 + 4^2}$

which you can solve for y

Edit: I missed the "Tangent at B" part of the question, and as pointed out by HallsOfIvy the line 3x-4y-21=0 does not pass through B, this solution finds the circle passing through A and B with tangent 3x-4y-21=0
• Jul 3rd 2009, 05:05 AM
HallsofIvy
Actually, the condition given, "passes through the point A(8,1) and B(7,1) and has , for its tangent at B, the line 3x-4y-21=0", is impossible. 3(7)- 4(1)- 21= 21- 4- 21= -4, not 0 so the line 3x- 4y-21= 0 does not even pass through B.

Tesla23's solution actually gives the circle passing through A and B and tangent to 3x- 4y- 21= 0 but not] at B.
• Jul 3rd 2009, 06:05 AM
helloying
Quote:

Originally Posted by HallsofIvy
Actually, the condition given, "passes through the point A(8,1) and B(7,1) and has , for its tangent at B, the line 3x-4y-21=0", is impossible. 3(7)- 4(1)- 21= 21- 4- 21= -4, not 0 so the line 3x- 4y-21= 0 does not even pass through B.

Tesla23's solution actually gives the circle passing through A and B and tangent to 3x- 4y- 21= 0 but not] at B.

Oh my it is not possible. But why did teachers set this qn for exam?is it a mistake or they just trying to test whether we are able to solve. is it possible for them to give another line that is possible and will pass through B?
• Jul 3rd 2009, 06:24 AM
Prove It
Quote:

Originally Posted by helloying
Oh my it is not possible. But why did teachers set this qn for exam?is it a mistake or they just trying to test whether we are able to solve. is it possible for them to give another line that is possible and will pass through B?