Question

If the line y=mx+c is a tangent to the circle x2+y2=a2,then the point of contact is​ ( The condition is not required here ,rather the point of contact is required ( the common solution of these equations)​

Answer 1

Answer:

(1) If (x1, y1 ) is a point outside the circle, then both the tangents are real. ... The condition for the line y = mx + c to be a tangent to the circle x2 + y2 = a2 is c2 = a2 (1+ m2). Then, c = ± √[9(1+16)] c = ±3 √17

Step-by-step explanation:

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Answer 2

Answer:

Instead of x we use b.

$\large{To \: Find:-}$

$The \: point \: of \: contact$

$\large{Solution:-}$

$The \: equation \: of \: the \: line \: is$

$y = mb + c$

$\longmapsto \bf {\mathbb{mx - y + c = 0}}$

$Here, a=m,b=−1,c=c$

$The \: equation \: of \: the \: circle \: is$

${b}^{2} + {y}^{2} = {a}^{2}$

$Line \: (i) \: will \: be \: a \: tangent \: \\ to \: the \: circle \: (ii), \: if \: \\ the \: length \: of \: perpendicular \\ from \: the \: centre \: (0,0) \: of \\ circle \: (ii) \: on \: line \: (i) \: = \\ the \: radius \: a.$

$\longmapsto{ \frac{ + }{ - } } \frac{m(0) - 0 + c}{ \sqrt{ {m}^{2} + 1} }$

$\longmapsto{c = \frac{ + }{ - } }a \sqrt{(1 + {m}^{2} )}$

$\longmapsto{ {c}^{2} } = {a}^{2} (1 + {m}^{2} )$

$Which \: is \: the \: required \: condition.$

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