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(a) The point $(−2, k)$ is on the circle $(x − 2)^2 + (y − 3)^2 = 65$ - Leaving Cert Mathematics - Question 3 - 2019

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(a)-The-point-$(−2,-k)$-is-on-the-circle-$(x-−-2)^2-+-(y-−-3)^2-=-65$-Leaving Cert Mathematics-Question 3-2019.png

(a) The point $(−2, k)$ is on the circle $(x − 2)^2 + (y − 3)^2 = 65$. Find the two possible values of $k$, where $k \in \mathbb{Z}$. (b) The circle $s$ is in the... show full transcript

Worked Solution & Example Answer:(a) The point $(−2, k)$ is on the circle $(x − 2)^2 + (y − 3)^2 = 65$ - Leaving Cert Mathematics - Question 3 - 2019

Step 1

Find the two possible values of $k$

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Answer

To find the values of kk, we substitute the point (2,k)(−2, k) into the equation of the circle:

(22)2+(k3)2=65(-2 - 2)^2 + (k - 3)^2 = 65

Calculating the left-hand side:

16+(k3)2=6516 + (k - 3)^2 = 65

This simplifies to:

(k3)2=49(k - 3)^2 = 49

We take the square root of both sides:

k3=7k - 3 = 7

or

k3=7k - 3 = -7

Solving for kk, we find:

  1. k=10k = 10
  2. k=4k = -4

Step 2

Find the equation of $s$

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Answer

The center of the circle ss will be at (g,g)(g, g), where gg is the radius of the circle since it touches both axes.
The distance from the center (g,g)(g, g) to the line 3x4y+6=03x - 4y + 6 = 0 should equal the radius.

First, we compute the perpendicular distance from the center to the line using:

d=3g4g+632+(4)2=g+65d = \frac{|3g - 4g + 6|}{\sqrt{3^2 + (-4)^2}} = \frac{|−g + 6|}{5}

Setting this equal to the radius gives:

g+6=g|−g + 6| = g

This simplifies to two equations:

  1. g+6=g−g + 6 = g, which gives g=3g = 3.
  2. g+6=g−g + 6 = −g, which does not provide a valid solution.

Thus, the center is (3,3)(3, 3) and the radius is 33. The equation of the circle can be expressed as:

(x3)2+(y3)2=32(x - 3)^2 + (y - 3)^2 = 3^2

This can be expanded to give:

x2+y26x6y+18=0x^2 + y^2 - 6x - 6y + 18 = 0

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