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Definition of Circle

The line r⃗...

Question

The line $\to r = \to a + t \to b$ touches the sphere ${\to r}^{2} - 2 \to r . \to c + h = 0, c^{2} > h$ at the point with position vector $\to a$ is

A

(\to a - \to b) . \to c = 0

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B

(\to a - \to c) . \to b = 0

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C

(\to b - \to c) . \to a = 0

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D

\to a . \to b + \to b . \to c + \to c . \to a = 0

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Solution

The correct option is B $(\to a - \to c) . \to b = 0$
The given line meets the given sphere at points for which
${(\to a + t {\to b}^{2})}^{2} - 2 (\to a + t \to b) . \to c + h = 0$
$\Rightarrow t^{2} {\to b}^{2} + 2 t (\to a . \to b - \to b . \to c) + {\to a}^{2} - 2 \to a . \to c + h = 0$ ...(1)
The line $\to r = \to a + t \to b$ touches the given sphere at $\to r = \to a,$ if the two values of $t$ obtained from (1) coincide with $t = 0$ .
$\Rightarrow {\to a}^{2} - 2 \to a . \to c + h = 0$ which shows $\to a$ lies on the sphere and $\to a . \to b - \to b . \to c = 0$
$\Rightarrow (\to a - \to c) . \to b = 0$

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