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Find the equation of the circle with its centre at (3, 2) and which touches to the line x+2y-4=0. |
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Answer» Solution :`ABS(bar(PC))=abs((ax_1+by_1+c)/SQRT(a^2+b^2))` = `(1 XX 3 + 2 xx 2 - 4)/sqrt(1^2+2^2) = 3/sqrt5` = radius `therefore` Equation of the CIRCLE is `(x-h)^2+(y-k)^2=a^2` or, `(x-3)^2+(y-2)^2=9/5` |
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