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Analytical Geometry

Hot goIITian

 Joined: 11 May 2011 Post: 163
17 Feb 2012 12:09:59 IST
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A circle is made to pass through the point (1,2), touching the straight lines 7x = y +5 an
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Analytical Geometry

A circle is made to pass through the point (1,2), touching the straight lines 7x = y +5 and x + y + 13 = 0. Find the area of quadrilateral formed by these tangents and the circle with smaller radius.

Cool goIITian

Joined: 21 Jul 2010
Posts: 88
17 Feb 2012 20:09:34 IST
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let the (g,f) represent the centre of the circle

given 7x = y +5 and x + y + 13 = 0 are tangents to the this circle and note that (1,2) lies on the line 7x = y +5

so, (f-2)/(g-1) = -1/7   {since the line joing the centre of the circle and (1,2) ie ,the radius is perpendicular to the line 7x = y +5}     .........(1)

now, radius of the circle = the perpendicular dist. between (g,f) and the lines 7x = y +5 , x + y + 13 = 0

ie, (7g-f-5)/root(50) and (g+f+13)/root2   { using the perpendicular dist. formula}

equating these two , (7g-f-5)/root(50) =    (g+f+13)/root2  .........(2)

solving (1) and (2) we will get 2 pairs of (g,f), take that pair which gives smaller radius  .

rest is easy.

Hot goIITian

Joined: 11 May 2011
Posts: 163
18 May 2012 14:20:44 IST
0 people liked this

@Prahlad thanks for solution

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