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Equation of a circle \mathrm{S(x, y)=0,(S(2,3)=16)} which touches the line \mathrm{3 x+4 y-7=0} at (1, 1) is given by

Option: 1

\mathrm{x^2+y^2+x+2 y-5=0}


Option: 2

\mathrm{x^2+y^2+2 x+2 y-6=0}


Option: 3

\mathrm{x^2+y^2+4 x-6 y=0}


Option: 4

none of these


Answers (1)

best_answer

Any circle which touches 3x + 4y – 7 = 0 at (1, 1) will be of the form

\mathrm{\mathrm{S}(\mathrm{x}, \mathrm{y}) \equiv(\mathrm{x}-1)^2+(\mathrm{y}-1)^2+\lambda(3 \mathrm{x}+4 \mathrm{y}-7)=0}

Since \mathrm{\mathrm{S}(2,3)=16 \Rightarrow \lambda=1}, so required circle will be

\mathrm{\mathrm{x}^2+\mathrm{y}^2+\mathrm{x}+2 \mathrm{y}-5=0}

Hence (a) is the correct answer.

Posted by

Devendra Khairwa

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