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The points $(0,4)$ and $(1,3)$ lie on a circle whose center is on the $x$-axis. What is the radius of the circle?

Guest Nov 21, 2017
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The points $(0,4)$ and $(1,3)$ lie on a circle whose center is on the $x$-axis.
What is the radius of the circle?

 

Let \(P_1 = (x_1,y_1) = (0,4)\)
Let \(P_2 = (x_2,y_2) = (1,3)\)
Let Center of the circle \(C = (x_c,y_c) = (\ ?,0)\)
Let radius of the circle = r

 

\(\begin{array}{|lrcll|} \hline & (x_c-x_1)^2 + (y_c-y_1)^2 &=& r^2 \quad & | \quad y_c = 0 \\ & (x_c-x_1)^2 + (0-y_1)^2 &=& r^2 \\ (1) & x_c^2-2x_cx_1+x_1^2 + y_1^2 &=& r^2 \\ \hline & (x_c-x_2)^2 + (y_c-y_2)^2 &=& r^2 \quad & | \quad y_c = 0 \\ & (x_c-x_2)^2 + (0-y_2)^2 &=& r^2 \\ (2) & x_c^2-2x_cx_2+x_2^2 + y_2^2 &=& r^2 \\ \hline \end{array} \)

 

\(\small{ \begin{array}{|lrcll|} \hline (1)-(2): & x_c^2-2x_cx_1+x_1^2 + y_1^2-(x_c^2-2x_cx_2+x_2^2 + y_2^2) &=& r^2-r^2 \\ & x_c^2-2x_cx_1+x_1^2 + y_1^2-x_c^2+2x_cx_2-(x_2^2 + y_2^2) &=& 0 \\ & -2x_cx_1+x_1^2 + y_1^2 +2x_cx_2-(x_2^2 + y_2^2) &=& 0 \\ & 2x_c(x_2-x_1)+x_1^2 + y_1^2 -(x_2^2 + y_2^2) &=& 0 \\ & 2x_c(x_2-x_1) &=& (x_2^2 + y_2^2)-(x_1^2 + y_1^2) \\ & \mathbf{ x_c } & \mathbf{=} & \mathbf{\dfrac{(x_2^2 + y_2^2)-(x_1^2 + y_1^2)} {2(x_2-x_1)}} \\ & && | \quad x_1 = 0 \quad y_1 = 4 \quad x_2 = 1 \quad y_2 = 3 \\ & x_c & = & \dfrac{(1^2 + 3^2)-(0^2 + 4^2)} {2(1-0)} \\ & x_c & = & \dfrac{10-16 } {2 } \\ & \mathbf{x_c} &\mathbf{ =} &\mathbf{ -3} \\\\ \hline (1) & x_c^2-2x_cx_1+x_1^2 + y_1^2 &=& r^2 \\ & (-3)^2-2(-3)(0)+0^2 + 4^2 &=& r^2 \\ & 9+16 &=& r^2 \\ & 25 &=& r^2 \\ & \mathbf{5} &\mathbf{ =}& \mathbf{r} \\ \hline \end{array} }\)

 

 

laugh

heureka  Nov 21, 2017
edited by heureka  Nov 21, 2017

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