Question

how to find the length of the tangent

Answer 1

Let the tangent drawn from the point P(x1,y1)P(x1,y1) meet the circle at the point TT as shown in the given diagram. The equation is given by

x2+y2+2gx+2fy+c=0 - - - (i)x2+y2+2gx+2fy+c=0 - - - (i)

Consider the triangle PTCPTC formed in this way is a right triangle, so according to the given diagram we have

|PT|2+|TC|2=|PC|2 - - - (ii)|PT|2+|TC|2=|PC|2 - - - (ii)

It is observed that |TC||TC| is the radius of the circle, so |TC|2=g2+f2−c|TC|2=g2+f2−c.

We also have

|PC|2=(x1−(−g))2+(y1−(−f))2=(x1+g)2+(y1+f)2|PC|2=(x1−(−g))2+(y1−(−f))2=(x1+g)2+(y1+f)2

Putting all these values in (ii), we get

|PT|2+g2+f2−c=(x1+g)2+(y1+f)2⇒|PT|2+g2+f2−c=x12+2gx1+g2+y12+2fy1+f2⇒|PT|2=x12+y12+2gx1+2fy1+c⇒|PT|=x12+y12+2gx1+2fy1+c−−−−−−−−−−−−−−−−−−−−−−√|PT|2+g2+f2−c=(x1+g)2+(y1+f)2⇒|PT|2+g2+f2−c=x12+2gx1+g2+y12+2fy1+f2⇒|PT|2=x12+y12+2gx1+2fy1+c⇒|PT|=x12+y12+2gx1+2fy1+c

This gives the length of the tangent from the point P(x1,y1)P(x1,y1) to the circle x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0.

Similarly, we can show that the PSPS is also of the same length.

Example: Find the length of the tangent from (12,−9)(12,−9) to the circle

3x2+3y2−7x+22y+9=03x2+3y2−7x+22y+9=0

Dividing the equation of the circle by 3, we get the standard form

x2+y2−73x+223y+3=0x2+y2−73x+223y+3=0

The required length of the tangent from (12,−9)(12,−9) is

(12)2+(−9)2−73(12)+223(−9)+3−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=144+81−28−66+3−−−−−−−−−−−−−−−−−−√=134

Answer 2

There are many formulae to find length of a tangent, which depends upon the figure given.

Suppose there are two tangents given and three points on the first tangent are A, P (in the middle) and B and three points on the other tangent are X, Y(the middle point) and Z.
Then the formula is PA x PB = YX x YZ.