Trying to find the right answer to this..

Question

Find the equation of a circle having centre at point C(–2, 3) and touching the straight line 3x – 4y – 2 = 0.

Solution

Centre of the required circle is C (-2, 3)
Let line AB with equation 3x - 4y - 2 = 0 touches the circle at point L
∴ p = r
where p, is length of perpendicular CL and r is the radius
$rightwards double arrow$ $open vertical bar fraction numerator 3 left parenthesis negative 2 right parenthesis minus 4 left parenthesis 3 right parenthesis minus 2 over denominator square root of left parenthesis 3 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction close vertical bar space equals space straight r space space space space rightwards double arrow space space straight r equals open vertical bar fraction numerator negative 20 over denominator 5 end fraction close vertical bar space equals space 4$
The equation of circle is:
$left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared space equals space straight r squared space or space left parenthesis straight x plus 2 right parenthesis squared space plus space left parenthesis straight y minus 3 right parenthesis squared space equals space left parenthesis 4 right parenthesis squared$
or $space space straight x squared plus straight y squared plus 4 straight x minus 6 straight y minus 3 equals 0$
Hence, the equation of circle is: $space space straight x squared plus straight y squared plus 4 straight x minus 6 straight y minus 3 equals 0$

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Binomial Theorem

Find the equation of a circle having centre at point C(–2, 3) and touching the straight line 3x – 4y – 2 = 0.

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