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Question -

Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y тАУ 1 = 0.



Answer -

It is given that weneed to find the equation of the circle with centre (3, 4) and touches thestraight line 5x + 12y тАУ 1 = 0.

We have a circle withcentre (3, 4) and having a radius┬а62/13.

We know that theequation of the circle with centre (p, q) and having radius тАШrтАЩ is given by: (xтАУ p)2┬а+ (y тАУ q)2┬а= r2

Now by substitutingthe values in the equation, we get

(x тАУ 3)2┬а+(y тАУ 4)2┬а= (62/13)2

x2┬атАУ6x + 9 + y2┬атАУ 8y + 16 = 3844/169

169x2┬а+169y2┬атАУ 1014x тАУ 1352y + 4225 = 3844

169x2┬а+169y2┬атАУ 1014x тАУ 1352y + 381 = 0

тИ┤ The equation of thecircle is 169x2┬а+ 169y2┬атАУ 1014x тАУ 1352y + 381 =0.

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