### Equations of straight lines and circles examples II

• Consider two circles $${C_{{\rm{ }}1}}{\kern 1pt} :{x^2} + {y^2} + 2x - 8y + 1 = 0$$ and $${C_{{\rm{ }}2}}{\kern 1pt} :{x^2} + {y^2} - 6x - 14y + 57 = 0$$.
(a) Find the coordinates of the centres and the radii of $${C_{{\rm{ }}1}}$$ and $${C_{{\rm{ }}2}}$$.
(b) Find the distance between the centres.
(c) Hence prove that $${C_{{\rm{ }}1}}$$ and $${C_{{\rm{ }}2}}$$ touch each other.
• (a) Coordinates of the centre of $${C_{{\rm{ }}1}}$$$$\begin{array}{l}\\ = ( - \frac{2}{2},{\rm{ }} - {\rm{ }}\frac{{ - {\rm{ }}8}}{2})\\ = {\underline{\underline {( - 1{\rm{ }},{\rm{ }}4)}} ^{}}\end{array}$$
Radius of $${C_{{\rm{ }}1}}$$$$\begin{array}{1}\\ = \sqrt {{{(\frac{2}{2})}^2} + {{(\frac{{ - {\rm{ }}8}}{2})}^2} - 1} \\ = {\underline{\underline 4} ^{}}\end{array}$$
Coordinates of the centre of $${C_{{\rm{ }}2}}$$$$\begin{array}{l}\\ = ( - \frac{{ - {\rm{ }}6}}{2},{\rm{ }} - {\rm{ }}\frac{{ - {\rm{ }}14}}{2})\\ = {\underline{\underline {(3{\rm{ }},{\rm{ }}7)}} ^{}}\end{array}$$
Radius of $${C_{{\rm{ }}2}}$$$$\begin{array}{1}\\ = \sqrt {{{(\frac{{ - {\rm{ }}6}}{2})}^2} + {{(\frac{{ - {\rm{ }}14}}{2})}^2} - 57} \\ = \underline{\underline {{\rm{ }}1{\rm{ }}}} _{}^{}\end{array}$$

(b) Distance between the centres$$\begin{array}{l} = \sqrt {{{[3 - ( - 1)]}^2} + {{(7 - 4)}^2}} \\ = {\sqrt {16 + 9} ^{}}\\ = {\underline{\underline 5} ^{}}\end{array}$$

(c) ∵ Radius of $${C_{{\rm{ }}1}}$$ + Radius of $${C_{{\rm{ }}2}}$$$$\begin{array}{l}\\ = 4 + 1\\ = {5^{}}\end{array}$$
= Distance between the centres
∴ $${C_{{\rm{ }}1}}$$ and $${C_{{\rm{ }}2}}$$ touch each other.

• G(-1, 0) is the centre of circle C. Straight line $$L{\kern 1pt} :y = - 2x + k$$ is the tangent to the circle at P.
(a) Express the coordinates of P in terms of k.
(b) If $$x + y + 4 = 0$$ passes through P, find the value of k.
(c) Find the equation of circle C.
• (a) Slope of L$$= - 2$$
∵ L is perpendicular to GP.
∴ Slope of GP$$= \frac{1}{2}$$
The equation of GP is
$$\begin{array}{1}y - 0 = \frac{1}{2}[x - ( - 1)]\\y = \frac{1}{2}(x + 1)\end{array}$$
$$\left\{ \begin{array}{l}y = - 2x + k\;................\;(1)\\y = \frac{1}{2}(x + 1)\;...............\;(2)\end{array} \right.$$
Substitute (2) into (1),
$$\begin{array}{1}\frac{1}{2}(x + 1) = - 2x + k\\x + 1 = - {\rm{ }}4x + 2k\\5x = 2k - 1\\x = \frac{{2k - 1}}{5}\end{array}$$
Substitute $$x = \frac{{2k - 1}}{5}$$ into (1),
$$\begin{array}{1}y = - 2(\frac{{2k - 1}}{5}) + k\\ = \frac{{k + 2}}{5}\end{array}$$
∴ The coordinates of P are $$(\frac{{2k - 1}}{5}{\rm{ }},{\rm{ }}\frac{{k + 2}}{5})$$.

$$\begin{array}{1} (b) \frac{{2k - 1}}{5} + \frac{{k + 2}}{5} + 4 = 0\\2k - 1 + k + 2 + 20 = 0\\3k = - 21\\k = \underline{\underline { - {\rm{ }}7}} \end{array}$$

(c) The coordinates of P are $$(\frac{{2( - 7) - 1}}{5}{\rm{ }},{\rm{ }}\frac{{ - {\rm{ }}7 + 2}}{5})$$, i.e. (-3, -1).
The equation of circle C is
$$\begin{array}{1}{[x - ( - 1)]^2} + {(y - 0)^2} = {[ - 1 - ( - 3)]^2} + {[0 - ( - 1)]^2}\\{(x + 1)^2} + {y^2} = 5\\{x^2} + 2x + 1 + {y^2} = 5\\{x^2} + {y^2} + 2x - 4 = 0\end{array}$$
• In the figure, a circle with its centre at C is the inscribed circle of ΔOAB, where P, Q and R are the points of contact.

(a) Find the coordinates of C.
(b) Find the equation of the circle.
• (a) Let (x, x) be the coordinates of C.
$$\begin{array}{1}BR = 6 - x\\BP = 6 - x\end{array}$$ (tangents from ext. pt.)
$$\begin{array}{1}AQ = 8 - x\\AP = 8 - x\end{array}$$ (tangents from ext. pt.)
In ΔOAB,
$$\begin{array}{1}A{B^2} = O{A^2} + O{B^2}\\ = {8^2} + {6^2}\\ = {100^{}}\\AB = {10^{}}\end{array}$$ (Pyth. theorem)
∴ $$\begin{array}{1}(6 - x) + (8 - x) = 10\\14 - 2x = {10^{}}\\x = {2^{}}\end{array}$$
∴ The coordinates of C are (2, 2).

(b) The equation of the circle is
$$\begin{array}{1}{(x - 2)^2} + {(y - 2)^2} = {2^2}\\{x^2} - 4x + 4 + {y^2} - 4y + 4 = 4\\{x^2} + {y^2} - 4x - 4y + 4 = 0\end{array}$$