First slide

Definition of a circle

Question

The equation of the circle which touches the axes of the coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is $x^{2} + y^{2} - 2 c x - 2 c y + c^{2} = 0,$
where $c,$ is

Moderate

A

$1, 6$
B

$2, 1$
C

$3, 6$
D

$6, 4$

Solution

The equation of a circle touching the coordinate axes is
$(x - c)^{2} + (y - c)^{2} = c^{2}$
This touches $\frac{x}{3} + \frac{y}{4} = 1,$ i.e. $4 x + 3 y - 12 = 0$
$∴ |\frac{4 c + 3 c - 12}{\sqrt{4^{2} + 3^{2}}}| = c \Rightarrow | 7 c - 12 | = 5 c \Rightarrow c = 6, 1$
Thus, the equation of the required circle is
$x^{2} + y^{2} - 2 c x - 2 c y + c^{2} = 0,$ where $c = 1, 6$

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