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: $5x - 12y + 10 = 0$ $\dots (i)$ $-5x + 12y + 16 = 0$ $\dots (ii)$ Since the slopes of both lines are eq

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(A) Given the two tangent lines to the circle: $5x - 12y + 10 = 0$ $\dots (i)$ $-5x + 12y + 16 = 0$ $\dots (ii)$ 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,let us re-evaluate the constant terms. The lines are $5x - 12y + 10 = 0$ and $5x - 12y - 16 = 0$. The distance $D = \frac{|10 - (-16)|}{\sqrt{5^2 + 12^2}} = \frac{26}{13} = 2$. Since the diameter $D = 2$,the radius $r = \frac{D}{2} = \frac{2}{2} = 1$.

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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