Solve the attached problem that involves circle and tangent

[chwala]

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

 

[fresh_42]

Looks ok. To solve part (b) you need to solve the quadratic function for $m.$

 

[chwala]

Looks ok. To solve part (b) you need to solve the quadratic function for $m.$

Thanks @fresh_42 , yap that part was easy for me...as indicated on ms...we make use of the discriminant $b^2-4ac=0$