Question

find the equation of the circle with centre at (3,1) and touching the line 8x-15y+25=0​

Answer 1

x² + y² - 6x - 2y + 6 = 0

Step-by-step explanation:

The given tangent line is

8x - 15y + 25 = 0 ..... (1)

Let the equation of the line perpendicular to (1) be

15x + 8y = k ..... (2)

Here, (2) no. straight line passes through the point (3, 1). Then

k = 45 + 8 = 53

Thus the normal line through the centre is

15x + 8y = 53 ..... (3)

We have

8x - 15y = - 25 ..... (1)

15x + 8y = 53 ..... (2)

Multiplying (1) by 15 and (2) by 8, we get

120x - 225y = - 375

120x + 64y = 424

On subtraction, we get

289y = 799

or, y = 47/17

Then x = 35/17

Thus the line (1) touches the circle at (35/17, 47/17)

If we draw a diameter through the centre, where the circle is touched by line (1), we will see that the centre of the circle is at (3, 1) and one end of the diameter is (35/17, 47/17)

Let (p, q) be the other end of the diameter.

Then,

(p + 35/17)/2 = 3, i.e., p = 67/17

(q + 47/17)/2 = 1, i.e., q = - 13/17

So the other end of the diameter is (67/17, - 13/17)

Hence the circle having two ends (35/17, 47/17) and (67/17, - 13/17) of a diameter is

(x - 35/17) (x - 67/17) + (y - 47/17) (y + 13/17) = 0

or, x² - 6x + 2345/189 + y ²- 2y - 611/289 = 0

or, x² + y² - 6x - 2y + 6 = 0