148 AKM.Y llf BMOMKTKY Cm. \ II. 



curve is the line through jpj, and perpendicular to the 

 tangent at P v Hence, to get the equation of the normal 

 at any given point, it is only necessary to write the equation 

 of the tangent at this point (Art. 81 >, ami then the equa- 

 tion of a line perpendicular to this tangent (Arts. 53. 

 and passing through the given point. Thus the equation 

 of the normal to the circle 



at the point PjS 



-.. (2) 



The coordinates G- and F of the center of the given 

 circle (1) satisfy equation (2); hence, every normal to a <> 

 passes through the center of the circle. 



If the center of the circle be at the origin, then Q- = 0, 

 .F = 0, and (7 = r 3 , and the equation (2) of the normal 

 becomes 



which reduces to x^y xyi= 0, an equation which could 

 have been derived for the circle x* + y 2 = r 2 in precisely t he 

 same way that equation (2) was derived from equation (1). 



EXERCISES 



1. Derive, by the secant method, the equation of the tangent to the 

 circle z* + y a = 2 rz, the point of contact being P l = (x v y,). 



2. Write the equation of the tangent to the circle : 

 (a) z 8 + y* = 25, the point of contact being (3, 4) ; 



(ft) z* + y* - 3 x + 10 y = 15, the point of contact being (4, -1 1) ; 

 (y) (z - 2)* + (y - 3)* = 10, the point of contact being (5, 4) ; 

 (5) 3 z a + 3 y 8 - 2 y - 4 x = 0, the point of contact being (0, 0). 



