Question

Find the radius of a circle with center at C(1,1), if a chord of length 6 is bisected at M(3,4)

Answer 1

Find the radius of a circle with center at C(1,1), if a chord of length 6 is bisected at M(3,4)

SOLUTION

In an x-y Cartesian coordinate system, the circle with centre coordinates $(x_{C},y_{C})$ and radius R is the set of all points (x, y) such that

$(x-x_{C})^{2}+(y-y_{C})^{2}=R^{2}$

In our case

$(x-1)^{2}+(y-1)^{2}=R^{2}$

Let the points A and B (the beginning and end of the chord) have coordinates:

$A(x_{A},y_{A}),\ B(x_{B},y_{B})$

Points A and B belong to the circle - which means that their coordinates satisfy the equation circle

\left\{\begin{array}[]{l}(x_{A}-1)^{2}+(y_{A}-1)^{2}=R^{2}\cr(x_{B}-1)^{2}+(y_{B}-1)^{2}=R^{2}\end{array}\right.

A chord AB of length 6:

$(x_{A}-x_{B})^{2}+(y_{A}-y_{B})^{2}=6^{2}$

point M - the middle of segment AB. This means that its coordinates are expressed in terms of the coordinates of points A and B as follows:

\left\{\begin{array}[]{l}x_{M}=\frac{x_{A}+x_{B}}{2}\cr y_{M}=\frac{y_{A}+y_{B}}{2}\end{array}\right.\Rightarrow\left\{\begin{array}[]{l}3=\frac{x_{A}+x_{B}}{2}\cr 4=\frac{y_{A}+y_{B}}{2}\end{array}\right.\Rightarrow\left\{\begin{array}[]{l}6=x_{A}+x_{B}\cr 8=y_{A}+y_{B}\end{array}\right.

So, we have a system of five equations for the five unknowns $R,\ x_{A},\ x_{B},\ y_{A},\ y_{B}:$

\left\{\begin{array}[]{l}(x_{A}-1)^{2}+(y_{A}-1)^{2}=R^{2}\cr(x_{B}-1)^{2}+(y_{B}-1)^{2}=R^{2}\cr(x_{A}-x_{B})^{2}+(y_{A}-y_{B})^{2}=6^{2}\cr 6=x_{A}+x_{B}\cr 8=y_{A}+y_{B}\end{array}\right.

$$(x_A - 1)^2 + (y_A - 1)^2 = (x_B - 1)^2 + (y_B - 1)^2 \rightleftharpoons$$ $$(x_A - 1)^2 - (x_B - 1)^2 = (y_B - 1)^2 - (y_A - 1)^2$$ $$((x_A - 1) - (x_B - 1))((x_A - 1) + (x_B - 1)) =$$ $$= ((y_B - 1) - (y_A - 1))((y_B - 1) + (y_A - 1))$$ $$(x_A - 1 - x_B + 1)(x_A - 1 + x_B - 1) = (y_B - 1 - y_A + 1)(y_B - 1 + y_A - 1) \rightleftarrows$$ $$(x_A - x_B)(x_A + x_B - 2) = (y_B - y_A)(y_B + y_A - 2)$$ $$\begin{cases} (x_A - x_B)(x_A + x_B - 2) = (y_B - y_A)(y_B + y_A - 2) \\ \quad (x_A - x_B)^2 + (y_A - y_B)^2 = 6^2 \\ \quad 6 = x_A + x_B \\ \quad 8 = y_A + y_B \end{cases}$$ $$\begin{cases} (x_A - x_B)(6 - 2) = (y_B - y_A)(8 - 2) \\ \quad (x_A - x_B)^2 + (y_A - y_B)^2 = 6^2 \\ \quad 6 = x_A + x_B \\ \quad 8 = y_A + y_B \end{cases}$$ $$\begin{cases} 4(x_A - x_B) = 6(y_B - y_A) \\ \quad (x_A - x_B)^2 + (y_A - y_B)^2 = 6^2 \\ \quad 6 = x_A + x_B \\ \quad 8 = y_A + y_B \end{cases}$$ $$\frac{4}{9}(x_A - x_B)^2 + (x_A - x_B)^2 = 6^2 \Rightarrow x_A - x_B = \sqrt{\frac{6^2 * 9}{9 + 4}} = \frac{3 * 6}{\sqrt{13}}$$ $$\begin{cases} \frac{3 * 6}{\sqrt{13}} = x_A - x_B \\ \quad 6 = x_A + x_B \end{cases} \Rightarrow x_A = 3 + \frac{9}{\sqrt{13}}$$ $$\frac{9}{4}(y_A - y_B)^2 + (y_A - y_B)^2 = 6^2 \Rightarrow y_A - y_B = \sqrt{\frac{6^2 * 4}{9 + 4}} = \frac{2 * 6}{\sqrt{13}}$$ $$\begin{cases} \frac{2 * 6}{\sqrt{13}} = y_A - y_B \\ \quad 8 = y_A + y_B \end{cases} \Rightarrow y_A = 4 + \frac{6}{\sqrt{13}}$$ $$\begin{aligned} R &= \sqrt{(x_A - 1)^2 + (y_A - 1)^2} = \sqrt{\left(4 + \frac{6}{\sqrt{13}} - 1\right)^2 + \left(3 + \frac{9}{\sqrt{13}} - 1\right)^2} = \\ &= \sqrt{\left(3 + \frac{6}{\sqrt{13}}\right)^2 + \left(2 + \frac{9}{\sqrt{13}}\right)^2} \approx 6.4783 \end{aligned}$$

ANSWER

$$R \approx 6.4783$$

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