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)?

Question

Evelyn Agard · Accepted Answer

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

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#

Paisley Johnson · Answer

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

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#.

Paisley Johnson · Answer

undefined To find the equations of the two tangents to the circle x^2 + y^2 - 2x - 6y + 6 = 0 that pass through the point P(-1,2), we can follow these steps:

1. Rewrite the equation of the circle in standard form by completing the square for both x and y terms:
   (x^2 - 2x) + (y^2 - 6y) = -6
   (x^2 - 2x + 1) + (y^2 - 6y + 9) = -6 + 1 + 9
   (x - 1)^2 + (y - 3)^2 = 4

2. The center of the circle is (1, 3) and the radius is 2.

3. Find the slope of the line passing through the center of the circle (1, 3) and the given point P(-1, 2):
   slope = (2 - 3) / (-1 - 1) = -1 / -2 = 1/2

4. The slope of the tangent line to a circle at a given point is the negative reciprocal of the slope of the radius at that point. Therefore, the slope of the tangent line is -2.

5. Use the point-slope form of a line to find the equations of the two tangent lines:
   -2 = (y - 2) / (x + 1)

6. Simplify the equation to find the equations of the two tangent lines:
   y - 2 = -2(x + 1)
   y - 2 = -2x - 2
   y = -2x

y - 2 = -2(x + 1)
   y - 2 = -2x - 2
   y = -2x - 4

Therefore, the equations of the two tangents to the circle x^2 + y^2 - 2x - 6y + 6 = 0 that pass through the point P(-1,2) are y = -2x and y = -2x - 4.