So the equation of the circle is \((x+1)^2+(y-3)^2=29.\)
Example 2 (non-calculator)
Circle \(C_1\) has equation \(x^2+y^2+4x-6y-12=0.\)
Circle \(C_2\) has centre \((1,-4).\)
The two circles have equal radii.
Find the equation of circle \(C_2.\)
Now, as circle \(C_2\) has the same radius, its equation is \((x-1)^2+(y+4)^2=25.\)
Example 3 (non-calculator)
Circle \(C_1\) has equation \((x-1)^2+(y+7)^2=16.\)
Circle \(C_2\) has equation \(x^2+y^2-12x-10y+12=0.\) (a) Write down the centres and radii of \(C_1\) and \(C_2\). (b) Show that \(C_1\) and \(C_2\) do not intersect.
The tangent and radius at P are perpendicular, so we use \(m_{\textsf{rad}}\tiny\ \normalsize m_{\textsf{tan}}=-1\) to obtain \(m_{\textsf{tan}}= \large\frac{4}{3}\normalsize.\)
Finally, we use the Nat 5 method to find the equation of the tangent:
Note that alternative forms of the final answer such as \(4x-3y=36\) or \(4x-3y-36=0\) are also acceptable.
Example 5 (non-calculator)
Show that the line with equation \(y=2x-3\) is a tangent to the circle with equation \(x^2+y^2-8x+11=0\) and find the coordinates of the point of contact.
We substitute the equation of the straight line into the equation of the circle and prove that there is only one solution:
There is only one solution, so the line is a tangent.
\(y=2x-3 = 2(2)-3 = 1\) so the point of contact is \((2,1).\)
Note: Instead of factorising the quadratic above and showing that it is a perfect square, we could have shown that the discriminant \(b^2-4ac=0.\)
Example 6 (non-calculator)
The circle with equation \(x^2+y^2-6x-14y+k=0\) meets the coordinate axes at exactly three points, none of which are the origin. What is the value of \(k\)?
\(2g=-6\) and \(2f=-14\) so the centre is \((-g,-f)=(3,7).\)
Because the question rules out the situation where the circle passes through the origin and then one point on each of the \(x-\) and \(y-\)axes, we know that the circle intersects one of the axes twice and the other once. This can only occur if the radius is 7 units.
Given that the equation \(x^2+y^2-2px-4py+3p+2=0\) represents a circle, determine the range of values of \(p.\)
Note: This answer requires the solution of a quadratic inequality, which is covered within the Polynomials and Quadratics topic. If you haven't covered that yet, skip this example and come back to it later.
Questions of this type depend on the fact that the radius must be a positive number, so we are going to solve the inequality \(g^2+f^2-c \gt 0.\)
But first, we need to define \(g,\) \(f\) and \(c\) in terms of \(p.\)
\(2g=-2p\) so \(g=-p\)
\(2f=-4p\) so \(f=-2p\)
\(c=3p+2\)
The working above gives us "critical values" of \(p=1\) and \(p=-\small\frac{2}{5}\normalsize.\)
In order to state the range of values of \(p\) we can consider the graph of \(y=(p-1)(5p+2).\)
The graph is a "happy" parabola with a minimum turning point between the roots, so the sections of the graph where \((p-1)(5p+2)\gt 0\) are where \(p\lt -\small\frac{2}{5}\normalsize\) or \(p\gt 1.\)
Example 8 (non-calculator)
SQA Higher Maths 2016 Paper 1 Q8
Show that the line with equation \(y=3x-5\) is a tangent to the circle with equation \(x^2+y^2+2x-4y-5=0\) and find the coordinates of the point of contact.
We substitute the equation of the straight line into the equation of the circle and prove that there is only one solution:
The tangent and radius at P are perpendicular, so we use \(m_{\textsf{rad}}\tiny\ \normalsize m_{\textsf{tan}}=-1\) to obtain \(m_{\textsf{tan}}= -3.\)
Finally, we use the Nat 5 method to find the equation of the tangent:
$$
\begin{eqnarray}
y-b &=& m(x-a)\\[6pt]
y-1 &=& -3(x+2)\\[6pt]
y-1 &=& -3x-6\\[6pt]
y &=& -3x-6+1\\[6pt]
y &=& -3x-5
\end{eqnarray}
$$
Example 10 (non-calculator)
SQA Higher Maths 2019 Paper 1 Q3
Circle \(C_1\) has equation \(x^2+y^2-6x-2y-26=0.\)
Circle \(C_2\) has centre \((4,-2).\)
The radius of \(C_2\) is equal to the radius of \(C_1\small.\)
Find the equation of circle \(C_2.\)