Question

Question #15b5e

Answer 1

The circle with the equation `(x-3/2)^2+(y+7/4)^2=(9/8)^2` or `64x^2+64y^2-192x+224y+259=0`

Explanation:

Since the distance from the center of the original circle to any midpoint in a chord spanning a constant central angle is also constant , the locus asked for is other circle with a shorter radius but the same center of the larger circle, as we can see in Fig. 1

The equation of the larger circle is
`4x^2+4y^2-12x+14y+1=0`
`x^2+y^2-3x+7/2y+1/4=0`

`-2x_0=-3` => `x_0=3/2`
`-2y_0=7/2` => `y_0=-7/4`
`x_0^2+y_0^2-R^2=1/4` => `R^2=9/4+49/16-1/4` => `R=sqrt(36+49-4)/4=sqrt(81)/4` => `R=9/4`
`-> (x-3/2)^2+(y+7/4)^2=(9/4)^2`

To find the shorter radius (`r`) consider the `triangle_(ABC)` whose sides are the chord `AB` and two radii `R` as shown below

Since the segment CM acts as a bisector of chord AB and therefore `A hat C M=(1/2) A hat C B=pi/3`, we can see that
`cos (pi/3)=r/R` => `r=Rcos (pi/3)=9/4*1/2` => `r=9/8`

So the equation of the shorter circle is

`->(x-3/2)^2+(y+7/4)^2=(9/8)^2`
Or
`x^2-3x+9/4+y^2+7/2y+49/16-81/64=0`
`x^2+y^2-3x+7/2y+(144+196-81)/64=0`
`x^2+y^2-3x+7/2y+259/64=0`
`->64x^2+64y^2-192x+224y+259=0`