Question

The radius of the circle touching the straight line 2x + y – 1 = 0 and 6x + 3y – 2 = 0 is

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Answer 1

Refer to the attached pic for the solution
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Answer 2

Answer:

The radius of the circle touching the straight line 2x + y – 1 = 0 and 6x + 3y – 2 = 0 is $\frac{1}{6\sqrt{5} }$

Step-by-step explanation:

Given:

Two straight lines as -

1. 2x + y – 1 = 0
2. 6x + 3y – 2 = 0

To Find:

The radius of the circle touching the straight line 2x + y – 1 = 0 and 6x + 3y – 2 = 0

Concept:

Circles

Solution:

The two lines are touching the circle just on the outside.

Thus, as per the equations, the lines are parallel as:

$\frac{2}{6}= \frac{1}{3}$ making the equations equal and thus, they are parallel

We know that lines,

$y = mx + C_{1}$

$y = mx + C_{2}$

The diameter between the lines = $\frac{|C_{2} -C_{1} |}{\sqrt{1+m^{2} } }$

Given lines,

2x+y-1=0

y = -2x +1....(1)

Equating equation 1 with the given standard formula of $y = mx + C_{1}$

$\frac{6x}{3} +\frac{3y}{3}-\frac{2}{3} = 0$

$2x + y - \frac{2}{3} =0$

$y = -2x+\frac{2}{3}$

Equating equation 1 with the given standard formula of $y = mx + C_{2}$

Therefore, $\frac{|\frac{2}{3}-1 |}{\sqrt{1 +(-2)^2} }$

Diameter is equal to $\frac{1}{3\sqrt{5} }$

As the radius is twice the diameter then,

Radius = Diamater/2 = $\frac{\frac{1}{3\sqrt{5} } }{2} =\frac{1}{6\sqrt{5} }$

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