Show that the line $7 y - x = 5$ touches the circles $x^{2} + y^{2} - 5 x + 5 y = 0$ and find the equation of the other parallel tangent.

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Solution

$x^{2} + y^{2} - 5 x + 5 y = 0$
centre of circle is $C (\frac{5}{2}, - \frac{5}{2})$
Radius of circle $= \sqrt{{(\frac{5}{2})}^{2} + {(- \frac{5}{2})}^{2} - 0} = \frac{5 \sqrt{2}}{2}$
Let the perendicular distance of line from centre $= p$
$7 y - x = 5 x - 7 y + 5 = 0$
$p = \frac{\frac{5}{2} - 7 (- \frac{5}{2}) + 5}{\sqrt{1 + 49}} = \frac{5 \sqrt{2}}{2} \Rightarrow p = r$
So the given line is tangent to circle
Slope of given line $= \frac{1}{7}$
So by using equation of tangent in slope form , equations of tangents are
$y = \frac{1}{7} x \pm \frac{5 \sqrt{2}}{2} \sqrt{1 + \frac{1}{49}} y = \frac{x \pm 25}{7} 7 y = x + 25 7 y - x = 25$

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