Let the tangents at the points A(4,−11) and B(8,−5) on the circle x2+y2−3x+10y−15=0, intersect at the point C. Then the radius of the circle, whose centre is C and the line joining A and B is its tangent, is equal to :
Solutions · 1
Official solution
Equation of AB :
\[y - ( - 5) = {{ - 5 + 11} \over {8 - 4}}(x - 8)\]
\[ \Rightarrow y + 5 = {3 \over 2}(x - 8)\]
\[ \Rightarrow 2y + 10 = 3x - 24\]
\[ \Rightarrow 3x - 2y - 34 = 0\] .... (1)
Also AB is cord of contact. And we know equation of cord of contact to a circle is T=0
\[ \Rightarrow xh + yk - {3 \over 2}(x + h) + 5(y + k) - 15 = 0\]
\[ \Rightarrow x\left( {h - {3 \over 2}} \right) + y(k + 5) + \left( { - {3 \over 2}h + 5k - 15} \right) = 0\] .... (2)
Comparing equation (1) and (2), we get
\[{{h - {3 \over 2}} \over 3} = {{k + 5} \over { - 2}} = {{ - {3 \over 2}h + 5k - 15} \over { - 34}}\]
\[\therefore\] −2h+3=3k+15
\[ \Rightarrow 3k + 2h = - 12\] ..... (3)
and
\[17(k + 5) = - {3 \over 2}h + 5k - 15\]
\[ \Rightarrow 12k + {3 \over 2}h - 100\]
\[ \Rightarrow 3h + 24k = - 200\] ..... (4)
Solving (3) and (4), we get
h=8 and \[k = - {{28} \over 3}\]
\[\therefore\] Point C is \[\left( {8, - {{28} \over 3}} \right)\]
Now radius of the circle whose centre is at C and tangent is AB is
\[ = \left| {{{3(8) - 2\left( { - {{28} \over 3}} \right) - 34} \over {\sqrt {{3^2} + {2^2}} }}} \right|\]
\[ = \left| {{{26} \over {3\sqrt {13} }}} \right|\]
\[ = {{2\sqrt {13} } \over 3}\]