If a circle passing through the point - 1 ,   0 touches y -axis at 0 ,   2 , then the x -intercept…

1. $\frac{5}{2}$
2. $5$
3. $\frac{3}{2}$
4. $3$

Solution

Let $\left(a, 2\right)$ be the centre of the circle

Equation of the circle is

${\left(x-a\right)}^2+{\left(y-2\right)}^2=a^{\mathrm{2 }}$

Put $\left(-\mathrm{1,0}\right)$ in the equation of circle ${\left(-1-a\right)}^2+{\left(-2\right)}^2=a^2$

$1+2a+a^2+4=a^2$

$2a=-5$

$a=-\frac{5}{2}$

Equation of circle is

${\left(x+\frac{5}{2}\right)}^2+{\left(y-2\right)}^2=\frac{25}{4}$

x intercept $=2\sqrt{g^2-c}$

$=2\sqrt{\frac{25}{4}-4}$

$=2\sqrt{\frac{9}{4}}$

$=3$

Asked in: JEE Main 2015 (11 Apr Online)

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