Equations of circle

Question 1,
the centre (O) of C is : ( 9/2 , -7/2 )
slope of OP: ( -7/2 - 0 ) / ( 9/2 - 2 ) = -7/5
slope of tangent: 5/7
By point-slope form, equation of tangent is:
(y-0) / (x-2) = 5/7
so, 5x -7y -10=0

Question 2,
(a) Sub (0,0) into C,
LHS = RHS
so C passes through the origin

(b) the centre (O) of C is : (-3,9)
slope of O and origin: 9/(-3) = -3
slope of tangent = 1/3
The equation of tangent is:
y/x = 1/3
x - 3y = 0

1 It is given that P(2,0) lies on circle C：x^2+y^2-9x+7y+14=0.Find theequation
of the tangent to C at P. Please help.
Sol
設切線：y－0=m(x－2)
mx－y－2m=0
x^2+y^2－9x+7y+14=0
(x^2－9x+20.25)+(y+2+7y+12.25)=-14+20.25+12.25
(x－4.5)^2+(y+3.5)^2=18.5
(4,5，－3.5)到mx－y－2m=0距離=√18.5
｜4.5m+3.5－2m｜/√(1+m^2)=√18.5
｜4.5m+3.5－2m｜=√18.5*/√(1+m^2)
｜5m+7｜=√74*/√(1+m^2)
25m^2－70m+49=74m^2+74
49m^2－70m+25=0
(7m－5)^2=0
m=5/7(重根)
切線：(5/7)x－y－10/7=0
5x－7y=10
or
x^2+y^2－9x+7y+14=0.
2x+2yy’－9+7y’=0
(2y+7)y’=9－2x
y’=(9－2x)/(2y+7)
y’(0，2)=(9－4)/(0+7)=5/7
切線：y－0=(5/7)(x－2)
7y=5x－10
5x－7y=10

2 Circle C：x^2+y^2+6x－18y=0 is given.
(a)Show that C passes through the origin.
Sol
0^2+0^2+6*0－18*0=0
So
Cpasses through the origin.
(b)Find the equation of the tangent to C at the origin.
Sol
設切線：y－0=m(x－0)
y－mx=0
x^2+y^2+6x－18y=0
(x^2+6x+9)+(y^2－18y+81)=90
(x+3)^2+(y－9)^2=90
｜9+3m｜/√(1+m^2)=√90
｜3+m｜=√10*√(1+m^2)
m^2+6m+9=10(1+m^2)
9m^2－6m+1=0
(3m－1)^2=0
m=1/3(重根)
切線：y－x/3=0
x－3y=0
or
x^2+y^2+6x－18y=0
2x+2yy’+6－18y’=0
(2y－18)y’=－2x－6
y’=(－2x－6)/(2y－18)=(x+3)/(9－y)
y’(0，0)=3/9=1/3
切線：y－0=(1/3)(x－0)
x－3y=0