Question

3. What are the intersection points of the line whose equation is y=3x +3 and the

circle whose equation is (x - 2)² + y2 = 25?

Answer 1

Points of intersection of these graphs are (-2, -3) and (0.6, 4.8).

Explanation:

Equation of the circle → (x - 2)² + y² = 25 ---------(1)

→ (x - 2)² + (y - 0)² = 5²

By comparing this equation with the standard equation of the circle,

(x - a)² + (y - b)²= r²

Here (a, b) is the center and r is the radius of the circle.

Therefore, center of the circle is (2, 0) and radius = 5 units

Second equation is a linear equation → y = 3x + 3 -------(2)

x-intercept of the equation → x = -1

y-intercept of the equation → y = 3

By graphing these equations we can get the point of intersections.

Solving these equations algebraically,

Substitute the value of y from equation (2) in the equation (1),

(x - 2)² + (3x + 3)² = 25

x² - 4x + 4 + 9x² + 18x + 9 = 25

10x² + 14x - 12 = 0

5x² + 7x - 6 = 0

x =
`(-7\pm √(7^2-4(5)(-6)))/(2(5))`

x =
`(-7\pm √(169))/(10)`

x =
`(-7\pm13)/(10)`

x = -2, 0.6

From equation (2),

y = -3, 4.8

Therefore, points of intersection of these graphs are (-2, -3) and (0.6, 4.8).