Question

Show that the line 3x–4y = 1 touches the circle x² +y² –2x+4y+1 = 0 find the coordinates of the point of contact.

Answer 1

Centre and radius of circle x² +y² –2x+4y+1 = 0

is (1,–2) and \(\sqrt{1+ 4 –1}\) = 2 respectively.

If the distance of a line 3x–4y = 1 from the centre (1,–2) is equal to radius then the line touches or it is tangent to a circle.

\(\left|\frac{3\times1-4(-2)-1}{\sqrt{3^2+4^2}}\right|\)  \(\left|\frac{ax_!+by_1+c}{\sqrt{a^2+b^2}}\right|=d\)

\(\left|\frac{3+8-1}{5}\right|\) = 2 

∴ line 3x–4y= 1 touches the circle.

Let point of contact be (x₁ ,y₁) then equation of tangent to a circle x² +y² –2x+4y+1=0 is

xx₁ +yy₁ –(x+x₁ )+2(y+y₁)+1 = 0_________________(1)

x(x₁ –1)+y(y₁ +2)–x₁ +2y₁ +1 = 0__________________(2)

and given line 3x–4y–1 = 0

(1) and (2) are idenfical then comparing (1) and (2) we get

\(\frac{x_1-1}{3}=\frac{y_1+2}{-4}=\frac{-x_1+2y_1+1}{-1}\)

–x₁ +1= –3x₁ +6y₁ +3 or 2x₁ –6y₁ –2=0

–y₁ –2=4x₁ –8y₁ –4 or 4x₁ –7y₁ –2=0

Solving these two equations of x₁ ,y₁ we get x₁ = -1/5 and y₁ = -2/5

∴ point of contact is \(\frac{-1}{5},\frac{-2}5\)