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# How to work out question on intersection of curve and equation of tangents to a circle?

I have 2 questions that i have answers to, but i have no idea how I got the answer, my notes don't make any sense. Please can someone help me answer them again? I would really appreciate it :)

1) Show that the line y = 3x + 1 crosses the curve y = x^2 + 3 at (1,4) and find the co-ordinates of the other point of intersection. (Answer - (2,7))

2) The equation of a circle is (x +2)^2 + (y - 3)^2 = 16, find the equations of the 4 tangents to the circle which are parallel to the co-ordinate axes. (Answer - (x=-6), (x=2), (y=-1), (y=7)

Please include answers step by step :)

Many many thanks :)

### 2 Answers

- 7 years agoFavorite Answer
1) Show that y = 3x + 1 and y = x^2 + 3 intersect at (1 , 4)

y = 3x + 1

x = 1

y = 3 * 1 + 1

y = 3 + 1

y = 4

y = x^2 + 3

x = 1

y = 1^2 + 3

y = 1 + 3

y = 4

(1 , 4) and (1 , 4) are the same point

Find the other point of intersection.

If a = b and a = c, then b = c

y = 3x + 1

y = x^2 + 3

3x + 1 = x^2 + 3

0 = x^2 - 3x + 3 - 1

0 = x^2 - 3x + 2

x = (3 +/- sqrt(9 - 8)) / 2

x = (3 +/- 1) / 2

x = 4/2 , 2/2

x = 2 , 1

y = x^2 + 3

y = 2^2 + 3

y = 4 + 3

y = 7

(2 , 7) is the other point of intersection

2)

The equation of a circle is (x + 2)^2 + (y - 3)^2 = 16

Find the equations of the 4 tangents that are parallel to the axes

The formula for a circle is (x - h)^2 + (y - k)^2 = r^2

Where (h , k) is the center

r is the radius

h = -2

k = 3

(-2 , 3) is the center

16 = r^2

4 = r

4 is the radius.

This means that the circle will pass through (-2 +/- 4 , 3) and (-2 , 3 +/- 4)

(-2 - 4 , 3)

(-2 + 4 , 3)

(-2 , 3 - 4)

(-2 , 3 + 4)

(-6 , 3)

(2 , 3)

(-2 , -1)

(-2 , 7)

x = -6 will be tangent to the circle at (-6 , 3)

x = 2 will be tangent to the circle at (2 , 3)

y = -1 will be tangent to the circle at (-2 , -1)

y = 7 will be tangent to the circle at (-2 , 7)

- MechEng2030Lv 77 years ago
1.)

To show that y = 3x + 1 crosses the curve y = x^2 + 3 at (1, 4) we have to confirm that when we plug x = 1 into either equation, we get y = 4.

Plugging in x = 1 into y = 3x + 1 => 3(1) + 1 = 4 (OK)

Plugging in x = 1 into y = x^2 + 3 => (1)^2 + 3 = 4 (OK)

To find the other point of intersection, do 3x + 1 = x^2 + 3

x^2 - 3x + 2 = 0

(x - 2)(x - 1) = 0

x = 1, x = 2

So we have another intersection at (2, 3(2) + 1) or (2, 7).

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2.)

The tangents which are parallel to the coordinate axis will be located at the points on the circle directly vertical of the center and directly horizontal of the center. Since the standard form of a circle is (x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center and r is the radius, we have a circle centered about (-2, 3) and of radius 4 in this case. Therefore the two points on the circle directly above and below the center will be (-2, 3 + 4) and (-2, 3 - 4) or (-2, 7) and (-2, -1). The two points on the circle directly to the left and right of the center will be (-2 + 4, 3) and (-2 - 4, 3) or (2, 3) and (-6, 3). Therefore the 4 tangents will be the y-coordinates of the points on the circle directly above and below the center or y = -1 and y = 7 and also the x-coordinates of the points on the circle directly to the left and to the right of the center or x = -6 and x = 2.