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Draw the circle $c: x^2 + y^2 = 25$ - Leaving Cert Mathematics - Question 3 - 2015

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Draw the circle $c: x^2 + y^2 = 25$. Show your scale on both axes. Verify, using algebra, that A(-4, 3) is on c. Find the equation of the circle with centre (-4, 3... show full transcript

Worked Solution & Example Answer:Draw the circle $c: x^2 + y^2 = 25$ - Leaving Cert Mathematics - Question 3 - 2015

Step 1

Draw the circle $c: x^2 + y^2 = 25$. Show your scale on both axes.

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Answer

To draw the circle defined by the equation x2+y2=25x^2 + y^2 = 25, we first note that the equation is of the form x2+y2=r2x^2 + y^2 = r^2, where rr is the radius of the circle. Here, the radius r=extsqrt(25)=5r = ext{sqrt}(25) = 5.

Steps to plot the circle:

  1. Draw the axes: Label the x-axis and y-axis.
  2. Determine the scale: Since our radius is 55, we can set a suitable scale on both axes, such as 1 unit = 1 grid line. This means you will mark points from -5 to 5 on both axes.
  3. Plot the center: The center of the circle is at the origin (0,0).
  4. Find and plot key points: From the center, move 5 units up to (0,5), down to (0,-5), left to (-5,0), and right to (5,0). These points help in sketching the circular curve.
  5. Draw the circle: Using these points, sketch a smooth circle.

Step 2

Verify, using algebra, that A(-4, 3) is on c.

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Answer

To verify that the point A(-4, 3) is on the circle, substitute x=−4x = -4 and y=3y = 3 into the circle's equation:

x2+y2=25x^2 + y^2 = 25 Substituting the values: (−4)2+(3)2=16+9=25(-4)^2 + (3)^2 = 16 + 9 = 25 Since the left-hand side equals the right-hand side (RHS), A(-4, 3) lies on the circle.

Step 3

Find the equation of the circle with centre (-4, 3) that passes through the point (3, 4).

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Answer

To find the equation of the circle with center (-4, 3) that passes through the point (3, 4), we first need to determine the radius:

  1. Calculate the radius: Using the distance formula, the radius rr is given by: r=extsqrt(x2−x1)2+(y2−y1)2r = ext{sqrt}{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2} With center (−4,3)(-4, 3) and point (3,4)(3, 4):

r=extsqrt(3−(−4))2+(4−3)2=extsqrt(3+4)2+12=extsqrt72+12=extsqrt49+1=extsqrt50r = ext{sqrt}{{(3 - (-4))}^2 + {(4 - 3)}^2} = ext{sqrt}{{(3 + 4)}^2 + {1}^2} = ext{sqrt}{{7}^2 + {1}^2} = ext{sqrt}{49 + 1} = ext{sqrt}{50}

  1. Write the equation of the circle: Using the standard form of the equation of a circle:
    (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 Where (h,k)(h, k) is the center. Thus, substituting (−4,3)(-4, 3) as the center and r=extsqrt50r = ext{sqrt}{50} gives: (x+4)2+(y−3)2=50(x + 4)^2 + (y - 3)^2 = 50

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