How do you write an equation of a circle whose center is at P(4,-5) and is tangent to the line -x+2y=1?

2 Answers
Jul 18, 2016

Answer:

#x^2+y^2-8x+10y-4=0#

Explanation:

Since the line is tangent to the circle, its distance from the center is equal to the radius, that's unknown:

d=r

#d=(|ax_0+by_0+c|)/sqrt(a^2+b^2)#

where a,b,c are, in this case:

#a=-1; b=2;c=-1#

and #(x_0,y_0)# is the center #P(4,-5)#

So

radius=#|-1(4)+2(-5)-1|/sqrt((-1)^2+2^2)#

#=|-4-10-1|/sqrt5#

#=15/sqrt5=(15sqrt(5))/5=3sqrt(5)#

Now you can substitute the values in the equation of the circle:

#(x-x_c)^2+(y-y_c)^2=r^2#

#(x-4)^2+(y+5)^2=(3sqrt(5))^2#

#x^2+y^2-8x+10y+16+25-45=0#

#x^2+y^2-8x+10y-4=0#

Jul 18, 2016

Answer:

#(x-4)^2+(y+5)^2- 45=0#

Explanation:

Giving a circle

#C->(x-x_0)^2+(y-y_0)^2-r^2=0#

and a line

#L->ax+by+c=0#

the tangency point #p_t={x_t,y_t}# is such that in #p_t# both declivities have the same value, or equivalently, their normal vectors are aligned.

Their normal vectors are

#vec n_C = {C_x,C_y} = {2(x_t-x_0),2(y_t-y_0)}# and
#vec n_L = {L_x,L_y} = {a,b}#

so at tangency

#vec n_C = lambda vec n_L#

The tangency point is obtained solving

#{ (2(x_t-x_0)+lambda a = 0), (2(y_t-y_0)+lambda b = 0), (a x_t + b y_t + c = 0) :}#

Here we know #x_0,y_0,a,b,c# and we are looking for #x_t,y_t,lambda#

but

#{ (x_t=-(a c - b^2 x_0 + a b y_0)/(a^2 + b^2)), (y_t = -(b c + a b x_0 - a^2 y_0)/(a^2 + b^2)), (lambda = (2 (c + a x_0 + b y_0))/(a^2 + b^2)) :}#

substituting the known values we obtain

#x_t=1,y_t=1,lambda=-6#

we know that

#r^2=(x_t-x_0)^2+(y_t-y_0)^2#

so the circle is

#C->(x-4)^2+(y+5)^2- 45=0#