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  • A line meets the co-ordinate axes in A and B. A circle is circumscribed about the triangle OAB. If <img alt="d_1 \: \: and \: \: d_2" src="https://entrancecorner.oncodecogs.com/gif.latex?d_1%20%5C%

A line meets the co-ordinate axes in A and B. A circle is circumscribed about the triangle OAB. If d_1 \: \: and \: \: d_2 are the distances of the tangent to the circle at the origin O from the points A and B respectively, then diameter of the circle is:

Option: 1

\frac{2 d_1 + d-2 }{2 }


Option: 2

\frac{d_1 +2 d_2 }{2 }


Option: 3

d_1 + d_2


Option: 4

\frac{d_1 d_2 }{d_1+ d_2 }


Answers (1)

best_answer

 

Condition of tangency -

\\Length\:of\:perpendicular\:from\:centre\:of\:circle\:(0,0)\:on\:the\:line\:y=mx+c\\is\:Radius\:of\:circle\:\\i.e.,\frac{\left|c\right|\:}{\sqrt{1+m^2}\:}=a

\mathbf{c=\pm a\sqrt{1+m^2}}

 

- wherein

If  y=mx+c  is a tangent to the circle x^{2}+y^{2}=a^{2}

 

 

Equation of circum circle of triangle OAB

        x^2 + y^2 -ax -by= 0

            Equation of tangent at origin ax + by = 0.

            d_1 = \frac{|a^2 |}{\sqrt{ a^2 + b^2 }} \: \: and \: \: d_2 = \frac{|b^2| }{\sqrt{ a^2 + b^2 }} \\\\ d_1 + d_2 = \sqrt{ a^2 + b^2 }

= diameter

 

 

Posted by

Suraj Bhandari

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