Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 07/Solution 1

We begin by implicitly differentiating the equation, which gives $$2x+2y{y}^{\prime }=2(x^2+y^2-2x)(2x+2y{y}^{\prime }-2).$$ Hence, when $x=0\text{and}y=1$ this becomes $$2y^{\prime }=2(1)(2y^{\prime }-2)=4y^{\prime }-4,$$ so $y^{\prime }=2$. The tangent line is therefore the line with slope $2$ passing through the point $(0,1)$, and hence is given by $$y=2x+1.$$