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
, where is the centre and is the radius.
From the information given, we can generate three equations in three unknowns, so that we can solve them simultaneously for and .
Substituting point A into the equation gives
.
Substituting point B into the equation gives
.
This circle's derivative is given by:
.
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 , we can rearrange this to read .
So the gradient of the tangent at point B is .
So at point the gradient is .
Substituting these values into the derivative gives
.
We can now solve for and .
So far we have:
and .
Therefore
.
We also know:
, so
.
Finally, we know that
.
Now we finally have enough information for the equation of the circle:
.
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:
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
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.