What is the max value?

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3 Answers
Apr 9, 2018

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...


My notebook...My notebook...

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...

Apr 9, 2018

PQ^2=14+5sqrt3

Explanation:

enter image source here
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:)
enter image source here
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

Apr 9, 2018

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