Math, asked by prateek2724, 10 months ago

The length of intercept on the straight line 3x + 4y -1 =0 by tge circle x^2 +y^2 -6x -6y -7 =0 is

Answers

Answered by eudora

Length of the intercept on the straight line by the circle is 6 units.

Step-by-step explanation:

Equation of the circle is x² + y² - 6x - 6y - 7 = 0

x² - 6x + 9 + y²- 6y + 9 = 7 + 9 + 9

(x - 3)²+ (y - 3)²= 25

Equation of the straight line is 3x + 4y - 1 = 0

Or y = $\frac{(1-3x)}{4}$

To find the point of intersection we will find the solution of the equations of straight line and the circle.

By placing the value of y in the equation of the circle.

(x - 3)²+ [ $(\frac{1-3x}{4})-3$ ] = 25

16(x - 3)²+ $(-3x-11)^{2}$ = 16×25

16(x² + 9 - 6x) + (9x²+121 + 66x) = 400

16x² + 144 - 96x + 9x² + 121 + 66x = 400

25x² - 30x + 265 = 400

25x² - 30x - 135 = 0

5x² - 6x - 27 = 0

x = $\frac{6\pm\sqrt{(-6)^{2}+(4\times 5\times 27)}}{10}$

x = $\frac{6\pm \sqrt{576}}{10}$

x = $\frac{6\pm 24}{10}$

x = -1.8, 3

y = $\frac{1+5.4}{4}, \frac{1-9}{4}$

y = 1.6, -2

Therefore, straight lines intersects the circle at two points (-1.8, 1.6) and (3, -2)

Distance between these points = $\sqrt{(y-y')^{2}+(x-x)^{2}}=\sqrt{(1.6+2)^{2}+(-1.8-3)^{2}}$

= $\sqrt{12.96+23.04}$

= $\sqrt{36}$

= 6 units

Learn more about the circle and tangents from https://brainly.in/question/9060

Answered by haswanthsai2005

Length of the intercept on the straight line by the circle is 6 units.

Step-by-step explanation: