If 5x - 12y + 10 = 0 and 12y - 5x + 16 = 0 ar | vedclass

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(A) Given the two tangent lines to the circle are: $5x - 12y + 10 = 0$ $\dots$ $(i)$ $-5x + 12y + 16 = 0$ $\dots$ $(ii)$ Since the slopes of both line

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(A) Given the two tangent lines to the circle are: $5x - 12y + 10 = 0$ $\dots$ $(i)$ $-5x + 12y + 16 = 0$ $\dots$ $(ii)$ Since the slopes of both lines are equal to $\frac{5}{12}$,the lines are parallel. The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Here,$A = 5, B = -12, C_1 = 10, C_2 = 16$. The distance between the tangents is the diameter of the circle: $d = \frac{|10 - 16|}{\sqrt{5^2 + (-12)^2}} = \frac{|-6|}{\sqrt{25 + 144}} = \frac{6}{13}$. Wait,re-evaluating the constant terms: Equation $(i): 5x - 12y + 10 = 0$ Equation $(ii): -5x + 12y + 16 = 0 \implies 5x - 12y - 16 = 0$ Distance $d = \frac{|10 - (-16)|}{\sqrt{5^2 + (-12)^2}} = \frac{26}{13} = 2$. Since the distance between parallel tangents is the diameter,$2r = 2$. Therefore,the radius $r = 1$.

If $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$ are two tangents to a circle,then the radius of the circle is

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