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If $5 x-12 y+10=0$ and $12 y-5 x+16=0$ are two tangents to a circle, then the radius of the circle is
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Given that, $5 x-12 y+10=0$ $\ldots$ (i)
and $\quad-5 x+12 y+16=0$ $\ldots$ (ii)
Slope of Eq. (i) $=\frac{5}{12}$
Slope of Eq. (ii) $=\frac{5}{12}$
Thus, Eqs. (i) and (ii) are parallel.
Therefore, distance between parallel lines $=$ diameters of the circle.
$\Rightarrow \frac{|10+16|}{\sqrt{25+144}}=2 \times$ Radius of the circle
$\Rightarrow \quad 2 \times$ Radius of circle $=\frac{26}{13}$
$\therefore \quad$ Radius of circle $=1$
and $\quad-5 x+12 y+16=0$ $\ldots$ (ii)
Slope of Eq. (i) $=\frac{5}{12}$
Slope of Eq. (ii) $=\frac{5}{12}$
Thus, Eqs. (i) and (ii) are parallel.
Therefore, distance between parallel lines $=$ diameters of the circle.
$\Rightarrow \frac{|10+16|}{\sqrt{25+144}}=2 \times$ Radius of the circle
$\Rightarrow \quad 2 \times$ Radius of circle $=\frac{26}{13}$
$\therefore \quad$ Radius of circle $=1$
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