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if y = 2x is a chord of the circle x² + y² - 10x =0

find the equation of the circle with this chord as diameter

Solution :

We have  y=2x is the chord of the circle
 
${\mathrm{x}}^2+{\mathrm{y}}^2-10\mathrm{x} = 0 .......\left(1\right)$

Let the chord be AB, so we will find the intersection point of line y=2x and circle (1).

$$\begin{gathered} \mathrm{Put} \mathrm{y} = 2\mathrm{x} \mathrm{in} \left(1\right), \mathrm{we} \mathrm{get} \\ {\mathrm{x}}^2 + {\left(2\mathrm{x}\right)}^2 - 10\mathrm{x} = 0 \\ \Rightarrow 5{\mathrm{x}}^2 - 10\mathrm{x} = 0 \\ \Rightarrow 5\mathrm{x}\left(\mathrm{x} - 2\right) = 0 \\ \Rightarrow \mathrm{x} = 0 \mathrm{or} \mathrm{x} = 2 \end{gathered}$$

Now, we get y = 0 or y = 4

Thus the coordinates of end points of the chord is A (0,0) and B(2,4).

Now consider these points A and B as the end points of diameter of  a circle.

The mid-point of AB is the centre of the required circle = $\left(\frac{0+2}{2}, \frac{0+4}{2}\right) = \left(1,2\right)$ = (h,k)

the distance AB is the length of the diameter AB.

$$\begin{gathered} \mathrm{AB} = \sqrt{{\left(2-0\right)}^2+{\left(4-0\right)}^2} = 2\sqrt{5} \\ \mathrm{So}, \mathrm{radius} , \mathrm{r} = \frac{\mathrm{AB}}{2} = \sqrt{5} \\ \mathrm{So}, \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{having} \mathrm{the} \mathrm{centre} \mathrm{at} \left(\mathrm{h},\mathrm{k}\right) \mathrm{and} \mathrm{radius} \mathrm{r} \mathrm{is} \\ {\left(\mathrm{x}-\mathrm{h}\right)}^2 + {\left(\mathrm{y}-\mathrm{k}\right)}^2 = {\mathrm{r}}^2 \\ \mathrm{Now}, \left(\mathrm{h},\mathrm{k}\right) = \left(1,2\right) \mathrm{and} \mathrm{r} = \sqrt{5} \\ \mathrm{So}, \mathrm{the} \mathrm{required} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{is}, \\ {\left(\mathrm{x}-1\right)}^2 + {\left(\mathrm{y}-2\right)}^2 = 5 \\ \Rightarrow {\mathrm{x}}^2 + {\mathrm{y}}^2 + 5 - 2\mathrm{x} - 4\mathrm{y} = 5 \\ \Rightarrow {\mathbf{x}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{y}}^{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{2}\mathbf{x}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{4}\mathbf{y}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0} \end{gathered}$$