Question

Show that the line 7x- 3y - 1 =0 touches the circle x^2+ y ^2+5x- 7y +4 =0 at point (1,2)​

Answer 1

Proof:

The given circle is

x² + y² + 5x - 7y + 4 = 0 ..... (1)

and the given straight line is

7x - 3y - 1 = 0 ..... (2)

From (2), we get

y = (7x - 1)/3

Substituting y = (7x - 1)/3 in (1), we get

x² + {(7x - 1)/3}² + 5x - 7 {(7x - 1)/3} + 4 = 0

or, x² + (49x² - 14x + 1)/9 + 5x - (49x - 7)/3 + 4 = 0

or, 9x² + 49x² - 14x + 1 + 45x - 147x +÷ 21 + 36 = 0

or, 58x² - 116x + 58 = 0

or, x² - 2x + 1 = 0

or, (x - 1)² = 0

or, x - 1 = 0

i.e., x = 1

Then y = (7 - 1)/3 = 6/3 = 2

i.e., y = 2

Therefore the given straight line touches the given circle at the point (1, 2).

Hence proved.