How do you find the equations of the two tangents to the circle #x^2 + y^2 - 2x - 6y + 6 = 0# which pass through the point P(-1,2)?

2 Answers
Harish Chandra Rajpoot · Ratnaker Mehta
Jul 18, 2018

#3x+4y-5=0\ \ # & # \ \ x+1=0#

Explanation:

Given equation of circle: #x^2+y^2-2x-6y+6=0# can re-written as

#(x-1)^2+(y-3)^2=4#

The above circle has center at #(1, 3)# & radius #2#

Let tangent passing through the point #(-1, 2)# be drawn at the point #(h, k)# on the circle: #x^2+y^2-2x-6y+6=0# then the point #(h, k)# will satisfy the equation of circle as follows

#h^2+k^2-2h-6k+6=0\ ...........(1)#

Now, the line joining points #(h, k)# & center #(1, 3)# will be perpendicular to the line joining the points #(h, k)# & #(-1, 2)# hence by condition of perpendicular lines we have

#\frac{k-3}{h-1}\times {k-2}/{h-(-1)}=-1#

#h^2+k^2-5k+5=0\ .............(2)#

Now, subtracting (2) from (1) we get

#h^2+k^2-2h-6k+6-(h^2+k^2-5k+5)=0-0#

#k=1-2h\ ...........(3)#

Substituting #k=1-2h# in (2), we get

#h^2+(1-2h)^2-5(1-2h)+5=0#

#5h^2+6h+1=0#

#(5h+1)(h+1)=0#

#h=-1/5, -1#

Substituting the values of #h# in (3), we get corresponding values of #k# as follows

#k=1-2(-1/5), k=1-2(-1)#

#k=7/5, 3#

Hence, the points at which tangents are drawn are #(-1/5, 7/5)# & #(-1, 3)# Thus there are two tangents drawn from external point #(-1, 2)#

Now, the equation of tangent line joining #(-1, 2)# & #(-1/5, 7/5)# is given as

#y-2=\frac{2-7/5}{-1-(-1/5)}(x-(-1))#

#3x+4y-5=0#

Similarly, the equation of tangent line joining #(-1, 2)# & #(-1, 3)# is given as

#y-2=\frac{2-3}{-1-(-1)}(x-(-1))#

#x+1=0#

hence, the equations of tangent lines drawn from the external point to the given circle are

#3x+4y-5=0# & #x+1=0#

Ratnaker Mehta
Jul 18, 2018

# x+1=0, and, 3x+4y-5=0#.

Explanation:

Let us solve the Problem using Geometry.

For this, we suppose that the point of contact of the required

tangent through #P(-1,2)# is #Q(h,k)# on the circle

# S : x^2+y^2-2x-6y+6=0#.

#S : (x-1)^2+(y-3)^2=2^2#, we see that the centre is #C(1,3).#

From Geometry, we know that, #CQ bot PQ#.

Hence, #"(the slope of CQ)"xx("the slope of "PQ)=-1#.

#:. {(k-3)/(h-1)}xx{(k-2)/(h+1)}=-1#.

#:. (k^2-5k+6)+(h^2-1)=0, #

# i.e., h^2+k^2-5k+5=0...................................(star^1)#.

Also, #Q in S. :. h^2+k^2-2h-6k+6=0...........(star^2)#.

#:. (star^1)-(star^2) rArr 2h+k-1=0, or, k=1-2h...(star^3)#.

Then, by #(star^1), h^2+(1-2h)^2-5(-2h)=0#.

#:. 5h^2+6h+1=0 rArr h=-1, or, h=-1/5#.

#:. k=1-2h=3, or, k=7/5#.

Thus, there are two tangents through #P(-1,2)# that touch #S#

at #Q_1(-1,3) and Q_2(-1/5,7/5)#.

Their eqns. are, # PQ_1:x=-1, and, #

#PQ_2:(y-2)/(7/5-2)=(x+1)/(-1/5+1), i.e., 3x+4y-5=0#.

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