How do you write the equation of the circle passing through (−2,1) and tangent to the line 3x − 2y = 6 at the point (4, 3)?

2 Answers
Dec 1, 2016

Answer:

#(x+2/7)^2+(y-41/7)^2=1300/49#

Explanation:

The circle equation is

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

Now it pass by the points #p_1=(-2,1)# and #p_2=(4,3)#

and also is tangent to the straight

#L->3x-2y-6=0#

So we have that for #C#

#2(x-x_0)dx+2(y-y_0)dy=0# or
#dy/dx=-(x-x_0)/(y-y_0)#

and also for #L#

#3dx-2dy=0# or
#dy/dx=3/2#

We also know that at #p_2#

#-(x_2-x_0)/(y_2-y_0)=3/2# so

#{((x_1-x_0)^2+(y_1-y_0)^2= r^2), ((x_2-x_0)^2+(y_2-y_0)^2=r^2), (-2(x_2-x_0)=3(y_2-y_0)):}#

Here the variables to find are #x_0,y_0,r#

Subtracting the first and the second we have a new reduced system

#{(x_1^2 - x_2^2 + y_1^2-y_2^2+2 x_0 (x_2-x_1) + 2 y_0( y_2 -y_1)=0),(-2(x_2-x_0)=3(y_2-y_0)):}# with solutions

#x_0=(3 x_1^2 - 3 x_2^2 + 3 (y_1 - y_2)^2 + 4 x_2 (y_2-y_1))/(6 (x_1 - x_2) - 4 y_1 + 4 y_2) = -2/7#

#y_0=((x_1 - x_2)^2 + y_1^2 + 3 (-x_1 + x_2) y_2 - y_2^2)/(3 (x_2 - x_1) + 2 (y_1 - y_2))=41/7#

finally

#r^2=(x_1-x_0)^2+(y_1-y_0)^2# giving

#r=(10 sqrt[13])/7#

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Dec 1, 2016

Another approach to the problem is to consider that the circle's center #p_0# is located at the intersection of two lines: #p_0 = L_1 nn L_2# where

#L_1-> abs(p-p_1)^2=abs(p-p_2)^2#
#L_2->p = p_2+lambda vec v#

where #p = (x,y),p_1=(-2,1)# and #p_2=(4,3)#

From #L->3x-2y-6=0# we get #vec v=(2,3)#

the parameter #lambda in RR#

Expanding #L_1# we get

#L_1 -> < p,p_2-p_1 > =1/2(abs(p_2)^2-abs(p_1)^2)#

substituting #p# from #L_2# into #L_1# we get

#< p_2+lambda_0 vec v,p_2-p_1 > = abs(p_2)^2- < p_2, p_1 > + lambda < vec v, p_2-p_1 ># so

#lambda_0 = -(abs(p_2)^2+abs(p_1)^2+2 < p_2, p_1 >)/(< vec v, p_2-p_1>)#

then

#p_0=p_2+lambda_0 vec v# and also

#r = abs(p_0-p_2)#