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Solution :

Equation of the circle is `x^(2)+y^(2)-2x+4y-4=0`. <br> Centre of the circle is C(1,-2) and radius is `sqrt((-1)^(2)+2^(2)-(-4))=3`. <br> Tangents are drawn to the circle from point P(2,3). <br> Equation of the line through point P having slope m is <br> `y-3=m(x-2)` <br> or `mx-y-2m+3=0` <br> If the line touches the circle , then the distance from the centre of the circle to this line will be radius of the circle. <br> `:. |(m(1)+2-2m+3)/(sqrt(m^(2)+1))|=3` <br> `implies (5-m)^(2)=9(m^(2)+1)` <br> `implies 8 m^(2)+10m-16=0` <br> `implies 4m^(2)+5m-8=0` <br> `implies m=(-5+- sqrt(153))/(8)` <br> Therefore, required equations of tangents are <br> `y=3=((-5+-sqrt(153))/(8))(x-2)`.