Circle

Given circle is
x2+y2−2x+4y+3=0
So, center=(−g,−f)=(‌

−2

2

,‌

4

2

)
=(−1,2)‌ and ‌c=3
Also, the equation of the polar of a point(x1,y1)w.r.t the circlex2+y2+2gx+2fy+c=0is
‌xx1+yy1+g(x+x1)+f(y+y1)+c=0
‌⇒xx1+yy1−(x+x1)+2(y+y1)+3=0
‌⇒xx1+yy1−x−x1+2y+2y1+3=0
‌⇒(x1−1)x+(y1+2)y−x1+2y1+3=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Given, polar equation is
2x−3y+1=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
By comparing the coefficients of two equations, we get
‌

x1−1

2

‌=‌

y1+2

−3

‌=‌

−x1+2y1+3

1

⇒‌‌‌

x1−1

2

‌=‌

y1+2

−3

‌ and ‌
‌

x1−1

2

‌=‌

−x1+2y1+3

1

⇒−3x1+3‌=2y1+4‌ and ‌x1−1
‌=−2x1+4y1+6
⇒3x1+2y1‌=−1‌ and ‌3x1−4y1−7=0
By solving these two equations, we getx1=‌

5

9

andy1=‌

−4

3

‌ Now, ‌3x1−y1‌=3(‌

5

9

)−(‌

−4

3

)
‌=‌

5

3

+‌

4

3

=‌

9

3

=3
∴3x1−y1‌=3