the length of the tangent from (-4,6) to the circle x²+y² -3=0
Answers
Step-by-step explanation:
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 lpength.
