# Problem on tangents drawn to a circle

I am solving Co-ordinate geometry by S.L. Loney. I am stuck on a problem on circles involving tangents and chords. I am not sure, if my approach is correct to solving this problem. Any inputs, tips that would lead me to correctly solve this problem would help!

Tangents are drawn to circle $$x^2+y^2=12$$ at the points where it is met by the circle $$x^2+y^2-5x+3y-2=0$$. Find the point of intersection of these tangents.

Solution(My attempt).

The two circles have a common chord. If $$(x_1,y_1)$$ be the required, the chord of contact of the tangents drawn to the circle $$x^2+y^2=12$$ is:

$$xx_1+yy_1=12$$

But, the chord of contact of the tangents drawn through $$(x_1,y_1)$$ to the circle $$x^2+y^2-5x+3y-2=0$$ is:

$$xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0\\ xx_{1}+yy_{1}-\frac{5}{2}(x+x_{1})+\frac{3}{2}(y+y_{1})-2=0\\ 2xx_{1}+2yy_{1}-5(x+x_{1})+3(y+y_{1})-4=0\\ x(2x_{1}-5)+y(2y_{1}+3)-5x_{1}+3y_{1}-4=0$$

I am comparing the above two equations and attempting to solve for $$x_{1},y_{1}$$. Am I thinking on the right lines?

• Whay are you looking at the chord of contact of the second circle? – amd Feb 16 '19 at 8:37

Let $$A(a,b)$$ be the intersection point, $$C(x_1,y_1)$$ and $$D(x_2,y_2)$$ be intersection points of our circles.

Thus, the equation of the line $$CD$$ (the radical axis of our circles) it's $$x^2+y^2-12-(x^2+y^2-5x+3y-2)=0$$ or $$5x-3y=10.$$ Id est, we got the following system. $$ax_1+by_1=12,$$ $$ax_2+by_2=12,$$ $$5x_1-3y_1=10$$ and $$5x_2-3y_2=10,$$ which gives $$a(x_1-x_2)+b(y_1-y_2)=0$$ and $$5(x_1-x_2)-3(y_1-y_2)=0,$$ which gives $$A(5t,-3t)$$ for some real $$t$$.

Now, easy to see that $$A$$ is placed in the fourth quadrant, which says $$t>0$$.

Let $$K$$ be an intersection point of lines $$AO$$ and $$5x-3y=10.$$

Thus, $$CK\perp AO$$ and $$AC\perp CO$$, which gives $$CO^2=OK\cdot AO$$ or $$12=\frac{|-10|}{\sqrt{5^2+(-3)^2}}\cdot\sqrt{(5t)^2+(-3t)^2}$$ or $$|t|=\frac{6}{5},$$ which gives $$t=\frac{6}{5}$$ and $$A\left(6,-\frac{18}{5}\right).$$

yes you are thinking right. But you can do it relatively short. Don't find the equation of chord of contact individually just use equation of family of circles S1-kS2=0.(k not equal to -1). But if you replace k=-1 then it will give the chord of contact(much like solving the two circles). in this case it is -5x+3y+10=0 and cmpare it with chord of contact for any circle xx1+yy1-12=0 to get x1= 6 y1=18/5. tell me if i am doing it wrong.

• Welcome to the site! Please use MathJax. – Toby Mak Feb 16 '19 at 7:16

I propose a slightly different (and I believe efficient) approach. Once you have the radical axis $$r: 5x-3y = 10$$ consider that the common point between the tangents will be on the line passing through origin (center of $$\gamma : x^2+y^2 = 12$$) perpendicular to $$r$$, i.e. $$s:3x+5y=0.$$ If $$AB$$ is the chord cut out by $$\gamma$$ on $$r$$, $$M$$ is its mid-point, and $$C$$ is the intersection point you are looking for, then $$\triangle OAC$$ is right-angled and $$AM$$ is its altitude relative to the hypothenuse. So, by first Euclid's Theorem we get $$\overline{OA}^2 = \overline{OM}\cdot \overline{OC},$$ that is $$\frac{\overline{OC}}{\overline{OM}}=\frac{12}{\overline{OM}^2}.$$ $$M$$'s coordinates are obtained right away by intersection $$r$$ and $$s$$. We get $$M\left(\frac{25}{17},-\frac{15}{17}\right)$$. Therefore $$\overline{OM}^2 =\frac{850}{289},$$ and $$\frac{\overline{OC}}{\overline{OM}} = \frac{1734}{425}.$$ Thus $$x_C = x_M \cdot \frac{\overline{OC}}{\overline{OM}}=6,$$ and, from the equation of $$s$$, $$y_C = -x_C\cdot\frac{3}{5}=-\frac{18}{5}.$$

As a supplement to the other answers, I offer the following way to find the intersection point of the tangent lines: it is the pole of the radical axis. So, as per the other answers, subtract the equation of one circle from the other to get the equation $$5x-3y-10=0$$ of the radical axis, on which the two intersection points lie. Using the method described here, we compute $$\pmatrix{1&0&0\\0&1&0\\0&0&-\frac1{12}}\pmatrix{5\\-3\\-10} = \pmatrix{5\\-3\\\frac56},$$ so the tangents intersect at $$\left({5\over5/6},{-3\over5/6}\right)=\left(6,-\frac{18}5\right)$$.