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Mathematics · Ch 6 — Circle

Tangents from an External Point to a Circle

6.3.3

Tangents from an External Point to a Circle

Tangents from an External Point to a Circle

Setting up. Let P(x1,y1)P(x_1,y_1) be a point in the plane, lying outside the circle x2+y2=a2x^2+y^2=a^2 (so it is not on the circle itself). Suppose a tangent line from PP has slope mm; by the point-slope form, its equation is

y−y1=m(x−x1),i.e.mx−y−mx1+y1=0y-y_1=m(x-x_1), \quad\text{i.e.}\quad mx-y-mx_1+y_1=0

Applying the tangency condition. For this line to actually touch the circle, the perpendicular distance from the centre O(0,0)O(0,0) to it must equal the radius aa:

∣y1−mx1∣1+m2=a\frac{|y_1-mx_1|}{\sqrt{1+m^2}}=a

Squaring both sides: (y1−mx1)2=a2(1+m2)(y_1-mx_1)^2=a^2(1+m^2). Expanding the left side and collecting every term on one side as a polynomial in mm:

(x12−a2)m2−2x1y1m+(y12−a2)=0(x_1^2-a^2)m^2-2x_1y_1m+(y_1^2-a^2)=0

This is a quadratic equation in mm — so it has (in general) two roots, m1m_1 and m2m_2, which are the slopes of the two tangent lines from PP.

From any point PP outside a circle (and in the same plane), exactly two tangents can be drawn to the circle.

Sum and product of the two slopes. Reading the coefficients of the quadratic directly:

m1+m2=2x1y1x12−a2,m1m2=y12−a2x12−a2m_1+m_2=\frac{2x_1y_1}{x_1^2-a^2}, \qquad m_1m_2=\frac{y_1^2-a^2}{x_1^2-a^2} …

Figure Fig.6.9Fig. 6.9 — two tangents from an external point

What this figure shows. A circle with an external point P(x1, y1) outside it, and two tangent lines drawn from P touching the circle at two distinct points -- illustrating that two tangents, with two different slopes m1 and m2, exist from any external point. …