Question

Find the equation of the circle which has its centre at the point (3,4)and touches the straight line 5x+12y=1.

Answer 1

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Answer 2

Answer:

$(x-3)^{2}+(y-4)^{2}=(\frac{62}{13})^{2}$

Step-by-step explanation:

In the question,

The centre of the circle is = (3, 4)

Straight line's equation is given by,

5x + 12y = 1

So,

The radius of the circle is given by the distance of the centre of the circle from the line just touching the circumference of the circle,

So,

Distance of the point (a, b) from the line (px + qy = c) is given by,

$Distance=\frac{|pa+qb-c|}{\sqrt{p^{2}+q^{2}}}$

So,

Distance is given by,

$Distance=\frac{|5(3)+12(4)-1|}{\sqrt{5^{2}+12^{2}}}=\frac{62}{13}$

Therefore, the Radius of the circle is 62/13.

Now,

The equation of the circle with the center (a, b) and radius 'r' is given by,

$(x-a)^{2}+(y-b)^{2}=r^{2}$

So,

The Equation of the circle is given as,

$(x-3)^{2}+(y-4)^{2}=(\frac{62}{13})^{2}$