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?

3 What are the intersection points of the line whose equation is y3x 3 and the circle whose equation is x 2 y2 25 class=

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Answer:

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

Step-by-step 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 = [tex]\frac{-7\pm \sqrt{7^2-4(5)(-6)}}{2(5)}[/tex]

x = [tex]\frac{-7\pm \sqrt{169}}{10}[/tex]

x = [tex]\frac{-7\pm13}{10}[/tex]

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).

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