Let the straight lines $5 x - 3 y + 15 = 0$ and $5 x + 3 y - 15 = 0$ form a triangle with the x-axis. Then, the radius of the circle circumscribing this triangle is

\frac{8}{5}

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\frac{17}{5}

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\frac{12}{5}

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\frac{16}{5}

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Solution

The correct option is A $\frac{17}{5}$

Let $(x, y)$ be cir cum center of triangle..
$O A^{2} = O B^{2} = O C^{2} = R^{2} x^{2} + {(y - 5)}^{2} = {(x + 3)}^{2} + y^{2} = {(x - 3)}^{2} + y^{2}$

$F r o m f i r s t t w o r e l a t i a n$
$- 10 y + 25 = 9 + 6 x 6 x + 10 y = 16 3 x + 5 y = 8 x = 0 y = \frac{8}{5} f r o m l a s t t w o r e l a t i o n w e g e t, \Rightarrow 12 x = 0 x = 0$

Therefore,
$R = \sqrt{{(x + 3)}^{2} + y^{2}} = \sqrt{9 + \frac{64}{25}} = \frac{\sqrt{289}}{5} = \frac{17}{5}$

Hence, this is the answer.

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Let the straight lines 5x−3y+15=0 and 5x+3y−15=0 form a triangle with the x-axis. Then, the radius of the circle circumscribing this triangle is

The correct option is A 175Let (x,y) be cir cum center of triangle..OA2=OB2=OC2=R2x2+(y−5)2=(x+3)2+y2=(x−3)2+y2Fromfirsttworelatian−10y+25=9+6x6x+10y=163x+5y=8x=0y=85fromlasttworelationweget,⇒12x=0x=0Therefore,R=√(x+3)2+y2=√9+6425=√2895=175Hence, this is the answer.

Let the straight lines $5 x - 3 y + 15 = 0$ and $5 x + 3 y - 15 = 0$ form a triangle with the x-axis. Then, the radius of the circle circumscribing this triangle is