Question 1

LEVEL 6

What is the equation of the following circle graph?

Select the correct answer from the list below:

A: x^2+y^2=7

B: (x+7)^2+(y+7)^2=7

C: x^2+y^2=49

D: x^2-y^2=7

 

CORRECT ANSWER: C: x^2+y^2=49

WORKED SOLUTION:

We can see that this is a circle, so must have an equation that looks like:

x^2+y^2=r^2
We can also see that it crosses the axis at 7, meaning that r=7. So:

x^2+y^2=7^2
x^2+y^2=49

Question 2

LEVEL 6

Which graph is given by the equations x^2+y^2=16?

Select the correct answer from the list below:

A:

B:

C:

D:

 

CORRECT ANSWER:   A

WORKED SOLUTION:

Because the equation is x^2+y^2=… we know the it has a circle. To decide which of the two circles it is, we need to find its radius.

Because the usual formula for a circle is:

x^2+y^2=r^2

And we have:

x^2+y^2=16

Then we can say:

r^2=16

And so:

r=\sqrt{16}
r=4

Because the radius of the circle is 4, and so must cross the axis at 4, the answer is A.

Question 3

LEVEL 8

Find the equation of the tangent to the point on the following circle:

Select the correct answer from the list below:

A: y =\frac{3}{4}x+\frac{3}{50}

B: y =-\frac{4}{3}x+\frac{3}{50}

C: y =-\frac{3}{4}x-\frac{16}{9}

D: y =-\frac{4}{3}x+\frac{50}{3}

 

CORRECT ANSWER:   D: y =-\frac{4}{3}x+\frac{50}{3}

WORKED SOLUTION:

To start, we need to find the gradient of the radius passing from the centre of the circle to the point on the circumference.

\textrm{Gradient of line} =\frac{\textrm {change in y}}{\textrm {change in x}}=\frac{6-0}{8-0}=\frac{6}{8}=\frac{3}{4}

Because we know that the radius meets the tangent at a right angle, then we need to take the negative reciprocal of our gradient to find the tangent’s gradient.

\textrm{Gradient of tangent} =-\frac{4}{3}

We also know that because the tangent is a straight line, it must have an equation in the form y=mx+c. But, we know the gradient, so we can put that in.

y =-\frac{4}{3}x+c

Now, all we need to do is find the value of c by putting in the values of x and y from the coordinate.
x=8
y=6

6 =-\frac{4}{3}\times8+c
6 =-\frac{32}{3}+c
6 +\frac{32}{3}=c
\frac{50}{3}=c

Putting this value of c, we get:

y =-\frac{4}{3}x+\frac{50}{3}

Question 4

LEVEL 8

Find the equation of the tangent to the point on the following circle:

Select the correct answer from the list below:

A: y =\frac{4}{3}x+\frac{25}{3}

B: y =\frac{1}{3}x+\frac{25}{3}

C: y =3 x+\frac{25}{3}

D: y =\frac{4}{3}x-\frac{25}{3}

 

CORRECT ANSWER:  A

WORKED SOLUTION:

To start, we need to find the gradient of the radius passing from the centre of the circle to the point on the circumference.

\textrm{Gradient of line} =\frac{\textrm {change in y}}{\textrm {change in x}}=\frac{3-0}{-4-0}=\frac{3}{-4} =-\frac{3}{4}

Because we know that the radius meets the tangent at a right angle, then we need to take the negative reciprocal of our gradient to find the tangent’s gradient.

\textrm{Gradient of tangent} =\frac{4}{3}

We also know that because the tangent is a straight line, it must have an equation in the form y=mx+c. But, we know the gradient, so we can put that in.

y =\frac{4}{3}x+c

Now, all we need to do is find the value of c by putting in the values of x and y from the coordinate.
x=-4
y=3

3 =\frac{4}{3}\times(-4)+c
3 =-\frac{16}{3}+c
3 +\frac{16}{3}=c
\frac{25}{3}=c

Putting this value of c, we get:

y =\frac{4}{3}x+\frac{25}{3}

Question 5

LEVEL 8

Find the tangent to the circle x^2+y^2=13^2 at the point (5,12)

Select the correct answer from the list below:

A: y =-\frac{5}{12}x+\frac{12}{5}

B: y =-\frac{5}{12}x+\frac{169}{12}

C: y =\frac{5}{12}x-\frac{8}{3}

D: y =-\frac{2}{7}x+\frac{132}{13}

 

CORRECT ANSWER:  B: y =-\frac{5}{12}x+\frac{169}{12}

WORKED SOLUTION:

It can be helpful to first draw a diagram. It is helpful to note that it doesn’t have to be drawn perfectly, just any circle with the centre at (0,0) and a tangent line in the correct quadrant.

To start, we need to find the gradient of the radius passing from the centre of the circle to the point on the circumference.

\textrm{Gradient of line} =\frac{\textrm {change in y}}{\textrm {change in x}}=\frac{12-0}{5-0}=\frac{12}{5}

Because we know that the radius meets the tangent at a right angle, then we need to take the negative reciprocal of our gradient to find the tangent’s gradient.

\textrm{Gradient of tangent} =-\frac{5}{12}

We also know that because the tangent is a straight line, it must have an equation in the form y=mx+c. But, we know the gradient, so we can put that in.

y =-\frac{5}{12}x+c

Now, all we need to do is find the value of c by putting in the values of x and y from the coordinate.
x=5
y=12

12 =-\frac{25}{12}+c
12 +\frac{25}{12}=c
\frac{144}{12} +\frac{25}{12}=c
\frac{169}{12}=c

Putting this value of c, we get:

y =-\frac{5}{12}x+\frac{169}{12}