Question -
Answer -
It is given that weneed to find the equation of the circle with centre (3, 4) and touches thestraight line 5x + 12y – 1 = 0.
We have a circle withcentre (3, 4) and having a radius 62/13.
We know that theequation of the circle with centre (p, q) and having radius ‘r’ is given by: (x– p)2 + (y – q)2 = r2
Now by substitutingthe values in the equation, we get
(x – 3)2 +(y – 4)2 = (62/13)2
x2 –6x + 9 + y2 – 8y + 16 = 3844/169
169x2 +169y2 – 1014x – 1352y + 4225 = 3844
169x2 +169y2 – 1014x – 1352y + 381 = 0
∴ The equation of thecircle is 169x2 + 169y2 – 1014x – 1352y + 381 =0.