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timurjin [86]
3 years ago
5

A line and a circle intersect at the points A and B. Use the equations below to find the coordinates of the points of intersecti

on A and B. y = x + 4 ( x − 3 ) 2 + ( y − 5 ) 2 = 34
Mathematics
1 answer:
Margaret [11]3 years ago
4 0

Answer:

<h2>A(-2, 2) and B(6, 10)</h2>

Step-by-step explanation:

Given the equation of a line y = x + 4 and equation of a circle as

( x − 3 )² + ( y − 5 )² = 34, if the line and the circle intersect at points A and B, to get this points, we will substitute the equation of the line into that of the circle as shown;

We will have to expand the equation of the circle first before making the substitute.

( x − 3 )² + ( y − 5 )² = 34

x²-6x+9+y²-10y+25 = 34

x²+y²-6x+-10y+34-34 = 0

x²+y²-6x+-10y = 0

Substituting y = x+ 4 into the resulting expression;

x²+(x+4)²-6x+-10y = 0

x²+x²+8x+16-6x+-10(x+4) = 0

x²+x²+8x+16-6x+-10x-40 = 0

2x²-8x-24 = 0

x²-4x-12 = 0

(x²-6x)+(2x-12) = 0

x(x-6)+2(x-6) = 0

x+2 = 0 and x-6 = 0

x = -2 and 6

when x = -2;

y = -2+4

y = 2

when x = 6

y = 6+4

y = 10

The coordinates of the point of intersection are A(-2, 2) and B(6, 10).

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Jet001 [13]

A there are no real zeros

using the discriminant b² - 4ac  to determine the nature of the zeros

for y = x² + 4x + 5 ( with a = 1, b = 4 and c = 5 )

• If b² - 4ac > 0 there are 2 real and distinct zeros

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b² - 4ac = 16 - 20 = - 4

Since discriminant < 0 there are no real zeros



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