Question

find the equation of circle which has center on the line y=2 and which passes through the points (2,0)and (4,0)?

Answer 1

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Solution

The circle has centre on line $$y=2$$

Let the centre be $$(x_1, y_1)$$

$$\Rightarrow y_1=2$$

circle passses through $$(2, 0)$$ and $$(4, 0)$$

We know $$OA=OB=R$$

$$PA=PB$$(perpendicular from centre bisects the chord)

But $$PA+PB=\sqrt{(4-2)^2+D^2}=2$$

$$2PA=2$$

$$\Rightarrow PA=1$$

$$\therefore$$ Coordinates of $$P=(2+1, 0)\Rightarrow (3, 0)$$

Since OP is parallel to y-axis.

centre lies along the line $$x=3$$

we also know it lies along the line $$y=2$$

$$\therefore$$ centre $$=(3, 2)$$

$$\Rightarrow$$ Radius $$=\sqrt{(3-2)^2+(2-0)^2}$$

$$=\sqrt{5}$$

Equation of circle is $$(x-3)^2+(y-2)^2=5$$.

hope it's helpful to you