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

Line 7x - 3y - 1 = 0 touches the circle x² + y² + 5x - 7y + 4 = 0 at the point (1, 2).

Step-by-step explanation:

Equation of the line is given as 7x - 3y - 1 = 0 ---------(1)

and the circle is x² + y² + 5x - 7y + 4 = 0 ------(2)

From equation (1),

y = $\frac{1}{3}(7x-1)$

Now we place the value of y in the equation of circle to get the common point  where the line and circle touch each other.

x² + [$\frac{1}{3}(7x-1)$]² + 5x - 7[

x² + $\frac{1}{9}(7x-1)^{2}$ + 5x -$\frac{7}{3}(7x-1)$ + 4 = 0

9x² + (49x² + 1 - 14x) + 45x - 21(7x - 1) + 4 = 0

9x² + 49x² + 1 - 14x + 45x - 147x + 21 + 4 = 0

58x² - 116x + 58 = 0

x² - 2x + 1 = 0

(x - 1)² = 0

x = 1

From equation 1,

y = 2

We get only one solution (1, 2) by the solving the equations that means line doesn't intersect the circle, it just touches the circle at (1, 2).

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