1. **Problem statement:** The circle C touches the y-axis at point A(0,3) and passes through point B(2,7). 2. **Find the equation of the perpendicular bisector of AB.** - The midpoint M of AB is $$M=\left(\frac{0+2}{2}, \frac{3+7}{2}\right) = (1, 5)$$. - The slope of AB is $$m_{AB} = \frac{7-3}{2-0} = \frac{4}{2} = 2$$. - The slope of the perpendicular bisector is the negative reciprocal: $$m_{\perp} = -\frac{1}{2}$$. - Using point-slope form with point M: $$y - 5 = -\frac{1}{2}(x - 1)$$ - Simplify: $$y = -\frac{1}{2}x + \frac{1}{2} + 5 = -\frac{1}{2}x + \frac{11}{2}$$ 3. **Find the equation of circle C.** - Since C touches the y-axis at A(0,3), the center must be at $(h,k)$ with $h$ equal to the radius $r$ because the distance from center to y-axis is $|h|=r$. - The circle passes through A(0,3), so: $$r = |h| = \text{distance from }(h,k) \text{ to } (0,3) = \sqrt{(h-0)^2 + (k-3)^2}$$ - Squaring both sides: $$r^2 = h^2 = (h)^2 + (k-3)^2 \implies (k-3)^2 = 0 \implies k=3$$ - So center is $(h,3)$ and radius $r=|h|$. - The circle passes through B(2,7), so: $$ (2 - h)^2 + (7 - 3)^2 = r^2 = h^2 $$ - Simplify: $$ (2 - h)^2 + 4^2 = h^2 $$ $$ (2 - h)^2 + 16 = h^2 $$ $$ 4 - 4h + h^2 + 16 = h^2 $$ $$ 20 - 4h = 0 $$ $$ 4h = 20 \implies h = 5 $$ - Therefore, center is $(5,3)$ and radius $r=5$. - Equation of circle C: $$ (x - 5)^2 + (y - 3)^2 = 25 $$ 4. **Show that the tangent to C at B has equation $3x - 4y + 22 = 0$.** - The radius vector at B is from center $(5,3)$ to B$(2,7)$: $$ \vec{r} = (2 - 5, 7 - 3) = (-3, 4) $$ - The tangent line at B is perpendicular to radius vector, so its slope is the negative reciprocal of slope of radius vector. - Slope of radius vector: $$ m_r = \frac{4}{-3} = -\frac{4}{3} $$ - Slope of tangent line: $$ m_t = \frac{3}{4} $$ - Equation of tangent line at B(2,7): $$ y - 7 = \frac{3}{4}(x - 2) $$ $$ 4(y - 7) = 3(x - 2) $$ $$ 4y - 28 = 3x - 6 $$ $$ 3x - 4y + 22 = 0 $$ - This matches the given equation, confirming the tangent line. **Final answers:** - a) Perpendicular bisector: $$y = -\frac{1}{2}x + \frac{11}{2}$$ - b) Circle equation: $$(x - 5)^2 + (y - 3)^2 = 25$$ - c) Tangent at B: $$3x - 4y + 22 = 0$$