Let y=x (in the closed half-plane x0) be the equation of a chord of the circle C_1 of diam…

Question

Let $y=x$ (in the closed half-plane $x\ge0$) be the equation of a chord of the circle $C_1$ of diameter $10$ passing through the origin. Let $C_2$ be another circle described on the given chord as its diameter. If the equation of the chord of $C_2$, which passes through the point $(2,3)$ and is farthest from the centre of $C_2$, is $x+ay+b=0$, then $a-b$ is equal to:

Accepted Answer

$C_1$ has radius $5$, passes through the origin, and lies in $x\ge0$, so its centre is $(5,0)$: $C_1:(x-5)^2+y^2=25$.

Intersecting with $y=x$: $(x-5)^2+x^2=25 \Rightarrow 2x^2-10x=0 \Rightarrow x=0,5$, giving points $(0,0)$ and $(5,5)$ --- the diameter of $C_2$.

Centre of $C_2 = \left(\dfrac52,\dfrac52\right)$. The chord of $C_2$ through $(2,3)$ farthest from the centre is perpendicular to the segment joining $(2,3)$ to the centre.

Slope of that segment: $\dfrac{5/2-3}{5/2-2}=\dfrac{-1/2}{1/2}=-1$, so the required chord has slope $1$:

$$y-3=1\cdot(x-2) \;\Rightarrow\; x-y+1=0.$$

Comparing with $x+ay+b=0$: $a=-1,\ b=1$, so $a-b=-2$.

Solutions · 1