Question

How do you find the tangent lines to this circle x2+y2–10x+8y−8=0 passing through the point A (-2,3)?

Answer 1

$\begin{aligned}&x^2+y^2-10x+8y-8=0\\ \\ \Longrightarrow\ \ &x^2+y^2-10x+8y+25+16-49=0\\ \\ \Longrightarrow\ \ &x^2-10x+25+y^2+8y+16=49\\ \\ \Longrightarrow\ \ &(x-5)^2+(y+4)^2=49\\ \\ \Longrightarrow\ \ &\sqrt{(x-5)^2+(y-(-4))^2}=7\ \ \ \ \ \longrightarrow\ \ \ (1)\end{aligned}$

From this, we get that the radius of the circle is 7 units and the center of the circle is (5, -4), where (x, y) is a point on the circle.

Distance between the center of the circle (5, -4) and the point (-2, 3),

$\Longrightarrow\ \sqrt{(5-(-2))^2+(-4-3)^2}\\ \\ \Longrightarrow\ \sqrt{7^2+(-7)^2}\\ \\ \Longrightarrow\ \sqrt{49+49}\\ \\ \Longrightarrow\ \sqrt{2 \times 49}\\ \\ \Longrightarrow\ \sqrt{2 \times 7^2}\\ \\ \Longrightarrow\ 7\sqrt{2}$

The length of the tangent through the point (-2, 3) to the circle can be found by taking the square root of the difference of the squares of the distance between points (5, -4) and (-2, 3) and the radius of the circle.

$\\ \\ \Longrightarrow\ \sqrt{(7\sqrt{2})^2-(7)^2}\\ \\ \Longrightarrow\ \sqrt{98-49}\\ \\ \Longrightarrow\ \sqrt{49}\\ \\ \Longrightarrow\ 7$

This is the length of the two tangents through the point (-2, 3) to the circle up to the circle.

Here we can find out that both the radius of the circle and the tangent to the  circle have same length, i.e., 7 units.

Let one tangent among the two meets the circle at the point (x, y).

$\begin{aligned}&\sqrt{(x-(-2))^2+(y-3)^2}=7\\ \\ \Longrightarrow\ \ &\sqrt{(x+2)^2+(y-3)^2}=7\ \ \ \ \ \longrightarrow\ \ \ (2)\end{aligned}$

$\begin{aligned}&(1)=(2)\\ \\ \Longrightarrow\ \ &\sqrt{(x-5)^2+(y+4)^2}=\sqrt{(x+2)^2+(y-3)^2}\\ \\ \Longrightarrow\ \ &(x-5)^2+(y+4)^2=(x+2)^2+(y-3)^2\\ \\ \Longrightarrow\ \ &x^2-10x+25+y^2+8y+16=x^2+4x+4+y^2-6y+9\\ \\ \Longrightarrow\ \ &4x+10x-6y-8y=25+16-4-9\\ \\ \Longrightarrow\ \ &14x-14y=28\\ \\ \Longrightarrow\ \ &14(x-y)=28\\ \\ \Longrightarrow\ \ &x-y=2\\ \\ \Longrightarrow\ \ &y=x-2\end{aligned}$

On again considering (2),

$\begin{aligned}&\sqrt{(x+2)^2+(y-3)^2}=7\\ \\ \Longrightarrow\ \ &\sqrt{(x+2)^2+(x-2-3)^2}=7\\ \\ \Longrightarrow\ \ &\sqrt{(x+2)^2+(x-5)^2}=7\\ \\ \Longrightarrow\ \ &(x+2)^2+(x-5)^2=49\\ \\ \Longrightarrow\ \ &x^2+4x+4+x^2-10x+25=49\\ \\ \Longrightarrow\ \ &2x^2-6x+29=49\\ \\ \Longrightarrow\ \ &2x^2-6x=20\\ \\ \Longrightarrow\ \ &x^2-3x=10\\ \\ \Longrightarrow\ \ &x^2-3x-10=0\\ \\ \Longrightarrow\ \ &x^2-5x+2x-10=0\\ \\ \Longrightarrow\ \ &x(x-5)+2(x-5)=0\end{aligned}$

$\begin{aligned}\\ \\ \Longrightarrow\ \ &(x+2)(x-5)=0\\ \\ \\ \therefore\ \ \ \ &x=-2\ \ \ ; \ \ \ x=5\\ \\ \therefore\ \ \ \ &y=-4\ \ \ ; \ \ \ y=3\ \ \ \ \ [\because\ y=x-2]\end{aligned}$

Hence the tangents meet the circle at points (-2, -4) and (5, 3) each.

Now we're going to find the equation of the tangent passing through the points (-2, 3) and (-2, -4).

Here this line is passing through points (-2, 3) and (-2, -4) which have same x coordinate. Thus the x coordinates of all points through which this line passes are always -2.

Hence the equation of this line is,

$\begin{aligned}&\Large x=-2\\ \\ \Large \Longrightarrow\ \ &\Large x+2=0\end{aligned}$

Now let's find the equation of the tangent passing through points (-2, 3) and (5, 3).

This line is passing through points (-2, 3) and (5, 3) whose y coordinates are the same. Thus the y coordinates of all points through which this line passes are always 3.

Hence,

$\begin{aligned}&\Large y=3\\ \\ \Large \Longrightarrow\ \ &\Large y-3=0\end{aligned}$