First slide

Analysis of two circles in circles

Question

P(a, b) is the point in the first quadrant. If the two circles which pass through P and touch both the coordinate axes cut at right angles, then

Moderate

A

$a^{2} - 6 a b + b^{2} = 0$
B

$a^{2} + 2 a b - b^{2} = 0$
C

$a^{2} - 4 a b + b^{2} = 0$
D

$a^{2} - 8 a b + b^{2} = 0$

Solution

Circle touches coordinate axes in first quadrant is ${(x - r)}^{2} + {(y - r)}^{2} = r^{2}$
$∴ {(a - r)}^{2} + {(b - r)}^{2} = r^{2} \Rightarrow r^{2} - 2 (a + b) r + (a^{2} + b^{2}) = 0$

Has two different real roots $r_{1}, r_{2}$ (radii)

$∴ {(C_{1} C_{2})}^{2} = r_{1}^{2} + r_{2}^{2}$

$\Rightarrow r_{1}^{2} + r_{2}^{2} - 4 r_{1} r_{2} = 0$

$\Rightarrow a^{2} + b^{2} - 4 a b = 0$

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