Question

What is the max value?

Answer 1

The answer is `(2)":- "color(red)(14+5sqrt3`

Check in image...

Explanation:

Put the coordinate `color(red)((0,-2))` on the equation and you will get the value greater than `color(red)(0`. It means the point `color(red)(p(0,-2)` is outside the circle.

Now check...

We determined the value of `bar"PQ"` by distance formula.

Not for JEE level, but a huge calculation. Will try to solve it with shorter method.

For any problem about handwriting , notify me.

:-)

Hope it helps...
Thank you...

Answer 2

`PQ^2=14+5sqrt3`

Explanation:

given a circle : `x^2+y^2-5x-y+5=0`,
rewrite the equation :
`=> (x-5/2)^2+(y-1/2)^2-25/4-1/4+5=0`
`=> (x-5/2)^2+(y-1/2)^2=3/2`
let `C(x,y) and r` be the center and the radius of the circle, respectively.
`=> C(x,y)=C(5/2,1/2), and r=sqrt(3/2)`
maximum `PQ=PC+CQ=PC+r`
`PQ=sqrt((5/2-0)^2+(1/2+2)^2)+sqrt(3/2)`
`=sqrt(50/4)+sqrt(3/2)`
`=(5sqrt2)/2+sqrt(3/2)`
`=(5+sqrt3)/(sqrt2)`
`=> PQ^2=((5+sqrt3)/(sqrt2))^2=14+5sqrt3`

Solution 2:)

`tana=(5/2)/(1/2+2)=1, => a=45^@`
`Q(x_q,y_q)=(x+rcos45+y+rsin45)`
`=(5/2+sqrt3/sqrt2*sqrt2/2, " "1/2+sqrt3/sqrt2*sqrt2/2)`
`=(5/2+sqrt3/2, ' " 1/2+sqrt3/2)=((5+sqrt3)/2, (1+sqrt3)/2)`
`=> PR=(5+sqrt3)/2-0=(5+sqrt3)/2`
`=> PQ=(PR)/cos45=(5+sqrt3)/2*2/sqrt2=(5+sqrt3)/sqrt2`
`=> PQ^2=((5+sqrt3)/sqrt2)^2=14+5sqrt3`

Answer 3

Given the equation of the circle `x^2+y^2-5x-y+5=0`.

The standard form of this equation :

`x^2-2*x*5/2+(5/2)^2+y^2-2*x*1/2+(1/2)^2=25/4+1/4-5=3/2`

`=> (x-5/2)^2+(y-1/2)^2=(sqrt(3/2))^2`

So coordinates of its center is `(5/2,1/2)` and its radius `r=sqrt(3/2)`

So coordinates of any point `(Q)` on the circle may be written parametrically as `(5/2+rcostheta,1/2+rsintheta)`

The coordinates of given point `(P)` is `(0,-2)`.

So

`PQ^2=(5/2+rcostheta)^2+(1/2+rsintheta+2)^2`

`=>PQ^2=(5/2+rcostheta)^2+(5/2+rsintheta)^2`

`=>PQ^2=25/4+r^2cos^2theta+25/4+r^2sin^2theta+2*5/2*rcostheta+2*5/2*sintheta`

`=>PQ^2=25/2+r^2(cos^2theta+sin^2theta)+5r(costheta+sintheta)`

`=>PQ^2=25/2+(sqrt(3/2))^2(cos^2theta+sin^2theta)+5sqrt(3/2)*sqrt2(1/sqrt2costheta+1/sqrt2sintheta)`

`=>PQ^2=25/2+3/2*1+5sqrt(3/2)*sqrt2(sin(pi/4)costheta+cos(pi/4)sintheta)`

`=>PQ^2=14+5sqrt3sin(theta+pi/4)`

`=>(PQ^2)_"max"=14+5sqrt3`, as maximum value of `sin(theta+pi/4)` is `1`