- #1

chwala

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- Homework Statement
- see attached.

- Relevant Equations
- circle equation.

Find Mark scheme here;

Find my approach here...more less the same with ms...if other methods are there kindly share...

part

**a (i)**

My approach is as follows;

##x^2+y^2-10x-14y+64=0 ##can also be expressed as

##(x-5)^2+(y-7)^2=10## The tangent line has the equation, ##y=mx+2## therefore it follows that,

##(x-5)^2+(mx+2-7)^2=10##

##(x-5)^2+(mx-5)^2=10##

##⇒(m^2+1)x^2-10(m+1)x+40=0##For

**part b**,

we need to find the co ordinates of point ##A## by

using ##(x-5)^2+(y-7)^2=10##, we know that ##y=3x+2##

therefore it follows that,

##(x-5)^2+(3x+2-7)^2-10=0##

##10x^2-40x+40=0##

##x^2-4x+4=0## giving us ##x=2, y=8##

##PA=\sqrt{(2-0)^2+(8-2)^2}=4+36=\sqrt 40=2\sqrt10##

##PB=\sqrt{(6-0)^2+(4-2)^2}=36+4=\sqrt 40=2\sqrt10##

Also to find the co ordinates at point ##B##,

we know that the tangent equation here is given by

##y=\frac{1}{3} x+2##

##x=3y-6## thus,

##(3y-11)^2+(y-7)^2-10=0##

##10y^2-80y+160=0## giving us ##y=4, x=6##

Therefore to find tan ##APB##, we shall use

##tan (A+B)=\dfrac {tan A + tan B}{1-tan A ⋅tan B}=\left[\dfrac {\dfrac{1}{2}+\dfrac{1}{2}}{1-\dfrac{1}{2} ⋅\dfrac{1}{2}}\right]=\left[\dfrac {1}{1-\dfrac{1}{4}}\right]=\left[\dfrac {1}{\dfrac{3}{4}}\right]=\left[\dfrac {4}{3}\right]##

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