Find centre of circle with equation of tangent given

Question

$(4, 1)$ is a point on one end of the diameter of a circle and the tangent through the other end of the diameter has equation $3x- y=1$. Determine the coordinates of the center of circle.

What got me about this question was that it didn't give me the circle center and ask me to find the equation of the tangent but instead is asking me to find the center with the tangent equation. I'm so confused as to how to work this out. Is it possible to find the coordinates of the equation? If so, could it be worked out the center by finding the midpoint of the two points? Can someone please tell me the process I need to go through to find the center of the circle, thanks.

Answer 1

There is a general expression derived in HCR's Web Papers for directly calculating the co-ordinates of foot of perpendicular say $(x', y')$ drawn from any point $(x_o, y_o)$ to the straight line $y=mx+c$ is given as $$(x', y')\equiv\left(\frac{x_o+m(y_o-c)}{1+m^2}, \frac{mx_o+m^2y_o+c}{1+m^2}\right)$$ Hence, the co-ordinate of the other end of diameter i.e. foot of perpendicular $(x', y')$ drawn from the point $(4, 1)$ to the line y=3x-1 are calculated as follows $$(x', y')\equiv\left(\frac{4+3(1-(-1))}{1+3^2}, \frac{3(4)+3^2(1)+(-1)}{1+3^2}\right)\equiv\left(\frac{10}{10}, \frac{20}{10} \right)\equiv\left(1, 2\right)$$ Hence, the co-ordinates of the center of the circle will be the mid-point of the line joining the end-points of the diameter $(4, 1)$ & $(1, 2)$ are calculated as follows $$\left(\frac{4+1}{2}, \frac{1+2}{2}\right)\equiv\left(\frac{5}{2}, \frac{3}{2}\right)$$

Answer 2

Refer to the given image. One as you can see, a diameter is always normal to a tangent!

So , we just need to find a equation of normal passing through $(4,1)$

Which is, $x+3y=k$ where $k$ is an indetermined constant. Now since the line pass through $(4,1)$ hence, $4+3=k$ or $k=7$

.Which gives equation of normal as $x+3y=7$ Now the point of intersection of this normal and tangent is the point $(a,b)$.

Which is $(1,2)$ , hence $(a,b)\equiv (1,2)$.

Also $(h,k)$ is mid point of $(1,2)$ , $(4,1)$ .

Which gives $(h,k)\equiv (\frac 52,\frac 32 ) $

Answer 3

Quite easy!

Since the point $(4,1)$ is tangent to $3x-y=1$; You find the equation of the tangent which is: $3y+x=3(1)+4=7$ So; $$ 3x-y=1 \tag{1}$$

$$ 3y+x=7 \tag{2} $$ solve simultaneously you have $x=1$, $y=2$ so point $(1,2)$ is the other end of the diameter so you have $(4,1)$ and $(1,2)$ as the two end of the diameter. Find the midpoint you have $(5/2, 3/2)$