1. **Problem Statement:** Given a circle with equation $$x^2 + y^2 = 25$$ and a line $$3x + 4y + 1 = 0$$, find: - (iii) A line parallel to the given line. - (iv) A line perpendicular to the tangent of the circle at a point where the tangent is perpendicular to the given line. 2. **Step 1: Find the slope of the given line** The line equation is $$3x + 4y + 1 = 0$$. Rewrite in slope-intercept form: $$4y = -3x - 1$$ $$y = -\frac{3}{4}x - \frac{1}{4}$$ So, slope $$m_1 = -\frac{3}{4}$$. 3. **Step 2: Find slope of line parallel to given line (iii)** Lines parallel have the same slope. So, slope of parallel line $$m = -\frac{3}{4}$$. 4. **Step 3: Find slope of line perpendicular to given line** If two lines are perpendicular, their slopes satisfy: $$m \times m_1 = -1$$ Substitute $$m_1 = -\frac{3}{4}$$: $$m \times \left(-\frac{3}{4}\right) = -1$$ $$m = \frac{4}{3}$$. 5. **Step 4: Differentiate the circle equation to find slope of tangent (iv)** Given: $$x^2 + y^2 = 25$$ Differentiate both sides w.r.t. $$x$$: $$2x + 2y \frac{dy}{dx} = 0$$ Solve for $$\frac{dy}{dx}$$: $$2y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = m_2 = -\frac{x}{y}$$ This is the slope of the tangent to the circle at point $$(x,y)$$. 6. **Step 5: Use condition that tangent is perpendicular to given line** Since tangent slope $$m_2$$ is perpendicular to line slope $$m_1$$, we have: $$m_2 = \frac{4}{3}$$ Substitute $$m_2 = -\frac{x}{y}$$: $$-\frac{x}{y} = \frac{4}{3}$$ Multiply both sides by $$y$$: $$-x = \frac{4}{3} y$$ Multiply both sides by 3: $$-3x = 4y$$ Or equivalently: $$-3y = 4x$$ 7. **Step 6: Find points on the circle satisfying the above** From $$-3y = 4x$$, express $$y$$ in terms of $$x$$: $$y = -\frac{4}{3} x$$ Substitute into circle equation: $$x^2 + y^2 = 25$$ $$x^2 + \left(-\frac{4}{3} x\right)^2 = 25$$ $$x^2 + \frac{16}{9} x^2 = 25$$ $$\frac{25}{9} x^2 = 25$$ Multiply both sides by 9: $$25 x^2 = 225$$ Divide both sides by 25: $$x^2 = 9$$ So, $$x = \pm 3$$ Find corresponding $$y$$: For $$x=3$$: $$y = -\frac{4}{3} \times 3 = -4$$ For $$x=-3$$: $$y = -\frac{4}{3} \times (-3) = 4$$ 8. **Step 7: Points of tangency** The points where the tangent is perpendicular to the given line are: $$(3, -4)$$ and $$(-3, 4)$$. **Final answers:** - (iii) A line parallel to $$3x + 4y + 1 = 0$$ has slope $$-\frac{3}{4}$$. - (iv) The slope of the line perpendicular to the tangent (which is perpendicular to the given line) is $$\frac{4}{3}$$. - The points on the circle where the tangent is perpendicular to the given line are $$(3, -4)$$ and $$(-3, 4)$$.