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A curve has equation x² + y² +12x = 64 A line has equation y = mx + 10 (a) (i) In the case that the line intersects the curve at two distinct points, show that (20m +12)²-144 (m² +1) > 0​

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User Wli
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Answer:

x² + (mx + 10)² + 12x - 64 = 0

x² + m²x² + 20mx + 100 + 12x - 64 = 0

(m² + 1)x² + (20m + 12)x + 36 = 0

In order for this equation to have two distinct solutions, the discriminant has to be greater than 0.

(20m + 12)² - 4(m² + 1)(36) > 0

(20m + 12)² - 144(m² + 1) > 0

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User Joel Worsham
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5 votes

Final answer:

The task involves finding where a line intersects a circle by using substitution to combine the equations. The resulting quadratic equation's discriminant provides the condition for two distinct intersection points, leading to the given inequality which demonstrates this condition.

Step-by-step explanation:

The student is working with two equations, one representing a circle and the other a line. To find where the line intersects the circle, they need to solve for x and y such that both equations are satisfied. Substituting the line's equation into the circle's equation gives an equation only in terms of x, which can be solved to find the points of intersection.

First, the line equation y = mx + 10 is substituted into the circle equation x² + y² +12x = 64, resulting in: x² + (mx + 10)² +12x = 64. Expanding and simplifying this leads to a quadratic equation in x. The condition for a quadratic equation ax² + bx + c = 0 to have two distinct real solutions is that its discriminant, b² - 4ac, must be greater than zero.

Applying this condition to the resulting quadratic of the intersection gives the inequality which can be expanded and simplified to the required form (20m +12)²-144 (m² +1) > 0. This inequality ensures that there are indeed two points of intersection between the line and the circle.

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User Chaos
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