# Thread: eqn of circle and tangent

1. ## 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

2. 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:

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

3. 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

4. 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.

5. 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?

6. 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?