The radius of the circle which touches $y -$ axis at $(0, 0)$ and passes through the point $(b, c)$ is:

\frac{b^{2} + c^{2}}{2 | b |}

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\frac{b^{2} + c^{2}}{2 | c |}

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\frac{b^{2} + c^{2}}{2}

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\frac{| b |}{2 (b^{2} + c^{2})}

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Solution

The correct option is A $\frac{b^{2} + c^{2}}{2 | b |}$
Given that,
The radius of circle $= ?$
touches the $y - a x i s$ then.
let $y - a x i s \to (a, 0)$ at $(0, 0)$
passes through the point $(b, c)$
then,
circle satisfies
$(x - a)^{2} + y^{2} = a^{2}$
$\Rightarrow x^{2} + a^{2} - 2 a x + y^{2} = a^{2}$
$\Rightarrow x^{2} - 2 a x + y^{2} = 0$
$\Rightarrow x^{2} + y^{2} = 2 a x$
$\Rightarrow a = \frac{x^{2} + y^{2}}{2 x}$
which is passes through $(b, c)$
then,
$\Rightarrow a = \frac{b^{2} + c^{2}}{2 b}$
Hence, the radius of circle is $\frac{b^{2} + c^{2}}{2 b}$

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If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

In a $△ A B C, 2 c a s i n (\frac{A - B + C}{2})$ is equal to

\frac{a^{2} - {(b - c)}^{2}}{{(a + c)}^{2} - b^{2}} + \frac{b^{2} - {(a - c)}^{2}}{{(a + b)}^{2} - c^{2}} + \frac{c^{2} - {(a - b)}^{2}}{{(b + c)}^{2} - a^{2}}

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

\frac{a^{2} - b^{2} - c^{2}}{(a - b) (a - c)} + \frac{b^{2} - c^{2} - a^{2}}{(b - c) (b - a)} + \frac{c^{2} - a^{2} - b^{2}}{(c - a) (c - b)}

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