How to find the centre of a circle if 2 points of the circle and a tangent to the circle is given?

Question

I am given a circle which has points (1,2) and (2,1). I am also given a tangent to the circle at another point. The Equation of the tangent is 4x-2y=9 .

I have assumed the centre of the circle to be (h,k). Then I calculated the distance between the points and the centre. I got the equation:

${(h-1)^2}+ {(k-2)^2} ={(h-2)^2}+{(k-1)^2} $

=>$ h=k$

Then I redrew the figure but this time the centre was (h,h) and the tangent touches the circle at point $(\frac{9+2y}{4} , y)$ (From the equation of the tangent)

After that I am stuck

How to find the Centre of the Circle with these informations?

Answer 1

The tangent line is $y=2x -\frac 92$ and the equation of the circle is $(x-h)^2+(y-h)^2 = (h-1)^2+ (h-2)^2$. Substituting $y$ to solve for the intersection point gives a quadratic in $x$

$$5x^2 - 6x(h+3)+15h +\frac{61}{4} =0$$ We need this to have only one solution, i.e. $$ 36(h+3)^2 - 20\left(15h+\frac{61}{4} \right) = 0 \\ \implies h= \frac 76 \pm \sqrt{\frac 56}$$

Answer 2

using the following equation $xx_p + y_py - α(x + x_p) - β(y + y_p) + F = 0$ you should be able to find the points of tangency as a function of h.From that point drop a perpendicular to (h,h) and if you measure its length using the formula $d =\frac{|Ax1 + By1 + C|} {(A^2 + B^2)^½}$ it must be equal to radius of the circle for which we already have an equation in terms of h, these two equations should suffice to find h.

Answer 3

You already proved the center is at $(h,h)$ for some $h$. Now, the radius is perpendicular to the tangent at the point of contact with the circle so that point (say $(p,q)$) is the point in the tangent that is closest to $(h,h)$. There are simple linear formulas for $p$ and $q$ in terms of $h$ and the parameters for the equation of the tangent, using those together with $$(p-h)^2 + (q-h)^2 = (h-1)^2 +(h-2)^2$$

gives you a quadratic equation for $h$.

Answer 4

The tangent line is also at the same distance from the center as the two points. Using standard point-point and point-line distance formulas produces the system of equations $$(1-h)^2+(2-k)^2 = (2-h)^2+(1-k)^2 = {(4h-2k-9)^2\over 4^2+2^2}.$$ The first equality reduces to $h=k$, and substituting that into either of the other two equations leaves a simple quadratic in either $h$ or $k$ to solve. We can see that the center must lie on the line $x=y$ even without this algebraic manipulation: it is somewhere on the bisector of $(1,2)$ and $(2,1)$, which by symmetry is $x=y$.