1.

Ihusitamon \( 15.3 \). Find the equation of the circle having radias 5 and which touches line \( 3 x+4 y-11=0 \) at point \( (1,2) \). Solution.

Answer»

let the equation of circle be-

\(x ^2+y ^2+2gx+2fy+c=0\)
centre of circle = (-g,-f)
radius of circle = \( \sqrt{f^2+g^2-c } = 5(given)\)
distance between the centre of circle and the line 3x+4y-11=0 is radius of the circle, as it touches the circle, i.e.5
therefore,
\( {|-3g-4f-11|\over \sqrt{3^2+4^2} }=5\)
→ -3g-4f-11=25
→ -3g-4f=36      ....(1)
now slope of normal to the circle at (1,2)
\(m = {2+f \over 1+g}\)
slope of the given line,  \(m' = {-3 \over 4}\)
therefore mm'=-1(normal and tangent to the circle at a point are always perpendicular)

→\( {2+f \over 1+g}={4\over 3}\)

→ 6+3f=4+4g

→ 4g-3f=2       ....(2)

multiply equation (1) and (2) by 4 and 3 respectively,

→   -12g-16f=144 ....(3)

→    12g-9f=6      ....(4)

add equation (3) and (4) we get,

→    -25f=150

→      f=\({150\over -25}=-6\)

substituting value of f in equation (1).

→   -3g+24=36

→   \(g= {36-24\over -3}=-4\)

thus the centre of circle is (4,6)

now, put the values of g and f in radius of circle equation \( \sqrt{f^2+g^2-c } = 5(given)\)

→      \( \sqrt{16+36-c } = 5\)

sqaring both sides we get,

52-c=25

→  c= 52-25= 27

hence the equation of circle is 

\(x^2+y^2-8x-12y+27=0\)


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