How do you find the equations of the two tangents to the circle x^2 + y^2 - 2x - 6y + 6 = 0x2+y22x6y+6=0 which pass through the point P(-1,2)?

2 Answers

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

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.