The mid point of the chord of the circle x2+y2−6x+4y−12=0 drawn parallel to the tangent at (-1, 1) and at a distance of one unit from the tangent is
Equation of circle is x2+y2−6x+4y−12=0 ∴2x+2yy′−6+4y′&=0 ⇒y′=‌
6−2x
2y+4
∴ Slope of tangent at (−1,1)=‌
dv
dx
|(−1,1) =‌
6−2(−1)
2(1)+4
=‌
8
6
=‌
4
3
∴ Equation of tangent is y−1=‌
4
3
(x+1) ⇒3y−3=4x+4 ⇒4x−3y+7=0 Let the equation of chord is 4x−3y+k=0 ∴‌‌‌
k−7
√16+9
=±1⇒k−7=±5 ⇒‌‌k=7±5⇒k=12,2 ∴ Equation of chord can be 4x−3y+12=0or 4x−3y+2=0 Now, Distance of both above lines from centre of circle are d1=|‌
12+6+12
5
|=6 and ‌‌d2=|‌
12+6+2
5
|=4 and radius of circle,r =√9+4+12=5 $ ∴ d_{2}∴ Equation of chord is 4x−3y+2=0
Now, slope of line 4x−3y+2=0 is ‌
4
3
∴ Slope of line CP=‌
−3
4
∴ Equation of CP is y+2=‌
−3
4
(x−3)
⇒‌‌4y+8=−3x+9⇒3x+4y=1
On solving 4x−3y+2=0 and 3x+4y=1, we get P(−‌