P is a point (a, b) in the first quadrant. If | vedclass

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Question

Equation of the two circles be $(x-r)^2+(y-r)^2=r^2$ (since it touches both the axes, $X$ coordinate of center $= Y$ coordinate of center $= r$ )

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Vedclass · Accepted Answer

$a^2 - 4ab + b^2 = 0$

Vedclass · Answer

Equation of the two circles be $(x-r)^2+(y-r)^2=r^2$ (since it touches both the axes, $X$ coordinate of center $= Y$ coordinate of center $= r$ )

i.e. $x^2+y^2-2 r x-2 r y+r^2=0$, where $r=r_1$ and $r_2$. Condition of orthogonality gives

$2 r _1 r _2+2 r _1 r _2= r _1^2+ r _2^2 \Rightarrow 4 r _1 r _2= r _1^2+ r _1^2 \ldots . .(1)$

Circle passes through $(a, b)$

$\Rightarrow a ^2+ b ^2-2 ra -2 r b + r ^2=0$

i.e. $r^2-2 r(a+b)+a^2+b^2=0$

$r _1+ r _2=2( a + b )$ and $r _1 r _2= a ^2+ b ^2\left( r _1, r _2ight.$ are roots of above equation)

Using these values in $(1)$, we get

$	herefore 4\left(a^2+b^2ight)=4(a+b)^2-2\left(a^2+b^2ight)$

i.e. $a^2-4 a b+b^2=0$

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

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