First slide

Definition of a circle

Question

The equation of the smallest circle passing from points $(1, 1)$ and $(2, 2)$ and always in the first quadrant, is

Moderate

A

$x^{2} + y^{2} - 4 x - 2 y + 4 = 0$
B

$x^{2} + y^{2} + 2 x + 4 y + 4 = 0$
C

$x^{2} + y^{2} - 3 x - 3 y + 4 = 0$
D

$x^{2} + y^{2} - 5 x - y + 4 = 0$

Solution

Let the equation of the required circle be
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
It passes through $(1, 1)$ and $(2, 2) .$
$\begin{array}{r} ∴ 2 + 2 g + 2 f + c = 0 \\ 8 + 4 g + 4 f + c = 0 \end{array}$
From (ii) and (iii), we get
$b + f = - 3$ and $c = 4$
Let $r$ be the radius of the circle. Then,
$\begin{array}{l} r^{2} = g^{2} + f^{2} - c \\ \Rightarrow r^{2} = g^{2} + f^{2} - 4 \\ \Rightarrow r^{2} = g^{2} + (3 + g)^{2} - 4 \\ \Rightarrow r^{2} = 2 g^{2} + 6 g + 5 \end{array}$ $[∵ g + f = - 3]$
$\Rightarrow r^{2} = 2 \{g^{2} + 3 g + \frac{5}{2}\} \Rightarrow r^{2} = 2 \{{(g + \frac{3}{2})}^{2} + \frac{1}{4}\}$
Clearly, $r$ is least when $g + \frac{3}{2} = 0 i.e. g = - \frac{3}{2}$
Substituting the values of $g, f$ and c in (i), we get
$x^{2} + y^{2} - 3 x - 3 y + 4 = 0$

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