A circle touching the x - axis at ( 3 ,   0 ) and making an intercept of length 8 on the y - axis…

1. $\left(3, 10\right)$
2. $\left(2, 3\right)$
3. $\left(3, 5\right)$
4. $\left(1, 5\right)$

Solution

Since circle touches the x-axis at $\left(3,\mathrm{ }0\right),$ hence centre can be taken as $\left(3,\mathrm{\alpha }\right)$ and radius $=\left|\mathrm{\alpha }\right|$
Since y-intercept is 8, by applying Pythagorean theorem
$3^2+4^2={\alpha }^2\Rightarrow \alpha =5,-5$
Hence two circles are possible
${\left(x-3\right)}^2+{\left(y-5\right)}^2=25$ and ${\left(x-3\right)}^2+{\left(y+5\right)}^2=25.$
Clearly $(3,\mathrm{ }10)$ satisfies the given equation of circle.

Asked in: JEE Main 2019 (12 Apr Shift 2)

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