Question

Circled one ....URGENT PLEASE

Answer 1

Answer:

Step-by-step explanation:

PA=PB

take square both side

PA^2=PB^2

now use distance

formula ,

{x-(a+b)}^2+{y-(b-a)}^2={x-(a-b)}^2+{y-(a+b)}^2

=>x^2+(a+b)^2-2x(a+b)+y^2+(b-a)^2-2y(b-a)y=x^2+(a-b)^2-2x(a-b)+y^2+(a+b)^2-2y(a+b)

=>2x(a-b)-2x(a+b)=2y(b-a)-2y(a+b)

=>2x{a-b-a-b}=2y{b-a-a-b}

=>2x(-2b)=2y(-2a)

=>bx=ay

hence proved

mark this as brainliest pl

Answer 2

Equal tangents,

BQ = BP

CP = CR

AQ = AR

Full answer is in attachment!

Hope it helps you!!