Question

Find the equation of a circle whose centre is C (2, –3) and which touches the straight line 3x – 4y – 3 = 0.

Solution

The centre of the circle is (2, -3)
$rightwards double arrow$ $left parenthesis straight h comma space straight k right parenthesis space left right arrow space left parenthesis 2 comma space minus 3 right parenthesis space space space rightwards double arrow space space straight h space equals space 2 comma space straight k equals negative 3$
The circle touches the straight line 3x - 4y - 3 = 0
$rightwards double arrow$ $space space straight r equals open vertical bar fraction numerator 3 straight h minus 4 straight k minus 3 over denominator square root of 9 plus 16 end root end fraction close vertical bar space space rightwards double arrow space straight r equals open vertical bar fraction numerator 3 left parenthesis 2 right parenthesis minus 4 left parenthesis negative 3 right parenthesis minus 3 over denominator 5 end fraction close vertical bar space rightwards double arrow space straight r space equals space open vertical bar fraction numerator 6 plus 12 minus 3 over denominator 5 end fraction close vertical bar equals 3$
Hence, the equation of the circle is $WiredFaculty$

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Binomial Theorem

Find the equation of a circle whose centre is C (2, –3) and which touches the straight line 3x – 4y – 3 = 0.

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