CircleHardBloom L3
Question
The equation of the circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is $x^{2} + y^{2} - 2rx - 2ry + r^{2} = 0$, then $r$ can be equal to:
Options
A.1
B.2
C.3
D.6
Solution
[Key Concept: A circle touching both axes with centre in the first quadrant has centre $(r,r)$; set its distance to the line $\frac{x}{3} + \frac{y}{4} = 1$ equal to $r$.]
Circles will be in circle and excentre circle of $\bigtriangleup ABC$ as shown
$$I \equiv \left( \frac{4(3)}{4 + 3 + 5},\frac{3(4)}{4 + 3 + 5} \right) \equiv (1,1)$$
$$\begin{matrix} I_{1} & \ \equiv \left( \frac{4(3)}{- 5 + 4 + 3},\frac{4(3)}{- 5 + 4 + 3} \right) \equiv (6,6) \\ r & \ = 1,6 \end{matrix}$$
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