length of tangent to a curve passing through another point

Question

Let $A$ be a point on the curve

$\mathcal{C} : x^2+y^2-2x-4=0$

If the tangent line to $\mathcal{C}$ at $A$ passes through $P(4,3)$, then what is the length of AP?

Please, include a general method of approaching similar kind of questions.

Answer 1

Bring the circle into standard form to find our center location and radius.

$$ ( h,k)=(1,0),\, R = \sqrt5 = CT $$

Distance of center to outside point squared

$$ PC^2 = (4-1)^2+ (3-0)^2= 18 $$

$$ PT^2= PC^2-5 = 18-5= 13 \rightarrow PT= \sqrt{13} $$

Answer 2

Welcome to MSE.Hint: first you have to find the center (o) of the curve which is a circle with equation:

$x^2-2x+y^2-4=(x-1)^2+(y-0)^2-5=0r^2$ or$(x-1)^2+(y-0)^2=5=r^2$

Where r is the radius of circle and O(-1, 0) is its center. In right triangle OAP we can write:

$AP^2=OP^2-r^2$

$OP^2=(4-1)^2+(3-0)^2=3^2+3^2=18$

$AP^2=18-5=13$ ⇒ $AP=\sqrt {13}$

Answer 3

$\mathcal{C}: x^2+y^2-2x-4=0$
$\Rightarrow x^2-2x+1-1+y^2-4=0$
$\Rightarrow (x-1)^2 +y^2=5=(\surd 5)^2$

Therefore, $\mathcal{C}$ is a circle with centre $O(1,0)$ and radius, $r=\surd 5$.

Let the tangent line to the circle $\mathcal{C}$ at the point $A$ and passing through $P(4,3)$ be $T$.

Then the perpendicular to $\mathcal{T}$ at point $A$ is a normal to the circle and passes through the centre, $O(1,0)$.

$AO$ is perpendicular to $\mathcal{T}$ (since, it is a normal)
$\Rightarrow AO$ is perpendicular to $AP$
$\Rightarrow (AO)^2 + (AP)^2 = (OP)^2$

$AO=r=\surd5$
$OP=$ distance between $(1,0)$ and $(4,3)= 3\surd2$
$AP$ is required.

Pre-requisite:
distance between two points $(x_1,y_1)$ and $(x_2,y_2)$
$=\surd[(x_1-x_2)^2+(y_1-y_2)^2]$

The answer will be clearer if your proceed with an image. It is better to draw the curves(if possible) while solving such questions!