CircleHardBloom L3
Question
A foot of the normal from the point $(4,3)$ to a circle is $(2,1)$ and a diameter of the circle has the equation $2x - y - 2 = 0$. Then the equation of the circle is:
Options
A.$x^{2} + y^{2} - 4y + 2 = 0$
B.$x^{2} + y^{2} - 4y + 1 = 0$
C.$x^{2} + y^{2} - 2x - 1 = 0$
D.$x^{2} + y^{2} - 2x + 1 = 0$
Solution
[Key Concept: Foot of the normal + a diameter fix the centre (midpoint) and radius of the circle.]
Diameter of the circle is given as
$$\begin{array}{r} 2x - y - 2 = 0\#(1) \end{array}$$
slope of $PN$ : $\frac{3 - 1}{4 - 2} = 1$
equation of normal through $PN$ is
$$\begin{matrix} y - 1 & & \ = (x - 2) \\ x - y - 1 & \ = 0 & \text{(2)} \end{matrix}$$
solving (1) and (2), centre is (1,0) Hence equation of the circle is
$$\begin{matrix} (x - 1)^{2} + y^{2} & \ = (2 - 1)^{2} + 1 \\ x^{2} + y^{2} - 2x - 1 & \ = 0 \end{matrix}$$
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