Let d y d x = a x - b y + a b x + c y + a , where a , b , c are constants. represent a circle passing…

Let dydx=ax-by+abx+cy+a, where a,b,c are constants. represent a circle passing through the point 2,5. Then the shortest distance of the point 11,6 from this circle is
  1. 10
  2. 8
  3. 7
  4. 5

Solution

Let equation of circle is

x2+y2+2gx+2fy+c=0

Differentiating above equation w.r.t x we get, 

dydx=-2x+2g2y+2f

Now, comparing with dydx=ax-by+abx+cy+a

We get, b=0,a=-2,c=2

-2g=-2g=1

Also, 2f=-2

So, f=-1

Now circle will be

x2+y2+2x-2y+c=0

its passes through 2,5

which will give c=-23

so circle will be x2+y2+2x-2y-23=0

centre C=-1,1

and radius 5

Now P is 11,6

So minimum distance of P from circle will be =11+12+6-12-5

=13-5=8

Asked in: JEE Main 2022 (27 Jun Shift 1)

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