Find the equations of the circles passing through the point `(-4,3)` and touching the lines `x+y=2` and `x-y=2`

Answer

`x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0`

Explanation

Let the equation of the required circle be
`x^(2)+y^(2)+2gx+2fy+c=0" "...(i)`
It passes through (-4, 3).
`25-8g+6f+c=0" "...(ii)`
Since, circle touches the line x + y - 2 = 0 and x - y - 2 = 0.
`therefore |(-g-f-2)/(sqrt2)|=|(-g+f-2)/(sqrt2)|=sqrt(g^(2)+f^(2)-c)" "...(iii)`
Now, ` |(-g-f-2)/(sqrt2)|=|(-g+f-2)/(sqrt2)|`
`rArr -g-f-2=pm(-g+f-2)`
`rArr -g-f-2=-g+f-2`
or `rArr -g-f-2=g-f+2`
`rArr f = 0 or g=-2`
Case I when f = 0
From Eq. (iii) we get
`|(-g-2)/(sqrt2)|=sqrt(g^(2)-c)`
`rArr (g+2)^(2)= 2(g^(2)-c)`
`rArr g^(2)-4g-4-2c=0" "...(iv)`
On putting f = 0 in Eq. (ii) we get
25 = 8g + c = 0 ...(v)
Eliminating c between Eqs. (iv) and (v),we get
`g^(2)-20g+46=0`
`rArr g = 10 pm3sqrt6 and c = 55pm24sqrt6`
On substracting the value of g, f and c in Eq.(i) we get
`x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0`
Case II when g = - 2
From Eq. (iii), we get
`rArr f^(2)=2(4+f^(2)-c)`
`rArr f^(2)-2c + 8 = 0 " "...(vi)`
On putting g = -2 in Eq. (ii), we get
c = -6f - 41
ON substituting c in Eq.(vi), we get
`f^(2)+12f+90=0`
this equation gives imageinary value of F.
Thus, there is no circle in this case.
Hence, the required equations of the circles are
`x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0`