Circle Line Tangent

1. **State the problem:** We have a circle $C$ with equation $$(x - 4)^2 + (y + 2)^2 = 13$$ and a line $L$ with equation $$y = mx + c$$ that passes through the point $(1, -1)$. (a) Express $c$ in terms of $m$. (b) Find the values of $m$ such that $L$ is tangent to $C$, and find the corresponding equation(s) of $L$. --- 2. **Part (a): Express $c$ in terms of $m$** Since $L$ passes through $(1, -1)$, substitute $x=1$ and $y=-1$ into $y = mx + c$: $$-1 = m(1) + c \implies c = -1 - m$$ So, $$c = -1 - m$$. --- 3. **Part (b): Find $m$ such that $L$ is tangent to $C$** Substitute $y = mx + c = mx - 1 - m$ into the circle equation: $$(x - 4)^2 + (y + 2)^2 = 13$$ Replace $y$: $$(x - 4)^2 + (mx - 1 - m + 2)^2 = 13$$ Simplify inside the second term: $$mx - 1 - m + 2 = mx + 1 - m$$ So the equation becomes: $$(x - 4)^2 + (mx + 1 - m)^2 = 13$$ Expand both squares: $$(x - 4)^2 = x^2 - 8x + 16$$ $$(mx + 1 - m)^2 = (mx)^2 + 2(mx)(1 - m) + (1 - m)^2 = m^2 x^2 + 2m x (1 - m) + (1 - m)^2$$ Sum: $$x^2 - 8x + 16 + m^2 x^2 + 2m x (1 - m) + (1 - m)^2 = 13$$ Group terms: $$(1 + m^2) x^2 + ig(-8 + 2m(1 - m)ig) x + ig(16 + (1 - m)^2 - 13ig) = 0$$ Simplify constant term: $$(1 - m)^2 = 1 - 2m + m^2$$ So constant term: $$16 + 1 - 2m + m^2 - 13 = 4 - 2m + m^2$$ The quadratic in $x$ is: $$ (1 + m^2) x^2 + (-8 + 2m - 2m^2) x + (4 - 2m + m^2) = 0$$ --- 4. **Condition for tangency:** The line is tangent to the circle if this quadratic has exactly one solution, i.e., its discriminant $D = 0$. Calculate discriminant: $$D = b^2 - 4ac$$ where $$a = 1 + m^2$$ $$b = -8 + 2m - 2m^2$$ $$c = 4 - 2m + m^2$$ Calculate $D$: $$D = (-8 + 2m - 2m^2)^2 - 4(1 + m^2)(4 - 2m + m^2)$$ Expand the first term: $$(-8 + 2m - 2m^2)^2 = ( -8 + 2m - 2m^2 )^2$$ Let’s write it as: $$(-8 + 2m - 2m^2)^2 = (-8)^2 + 2 imes (-8)(2m - 2m^2) + (2m - 2m^2)^2$$ Calculate stepwise: $$64 + 2 imes (-8)(2m - 2m^2) + (2m - 2m^2)^2$$ $$= 64 - 32m + 32 m^2 + (2m - 2m^2)^2$$ Now expand $(2m - 2m^2)^2$: $$= 4m^2 - 8 m^3 + 4 m^4$$ Sum all: $$64 - 32 m + 32 m^2 + 4 m^2 - 8 m^3 + 4 m^4 = 64 - 32 m + 36 m^2 - 8 m^3 + 4 m^4$$ Now expand the second term: $$4(1 + m^2)(4 - 2m + m^2) = 4 imes (4 - 2m + m^2 + 4 m^2 - 2 m^3 + m^4)$$ First expand inside: $$(1)(4 - 2m + m^2) + m^2 (4 - 2m + m^2) = 4 - 2m + m^2 + 4 m^2 - 2 m^3 + m^4 = 4 - 2 m + 5 m^2 - 2 m^3 + m^4$$ Multiply by 4: $$16 - 8 m + 20 m^2 - 8 m^3 + 4 m^4$$ --- 5. **Calculate $D$:** $$D = (64 - 32 m + 36 m^2 - 8 m^3 + 4 m^4) - (16 - 8 m + 20 m^2 - 8 m^3 + 4 m^4)$$ Simplify: $$D = 64 - 32 m + 36 m^2 - 8 m^3 + 4 m^4 - 16 + 8 m - 20 m^2 + 8 m^3 - 4 m^4$$ Combine like terms: Constants: $$64 - 16 = 48$$ $m$ terms: $$-32 m + 8 m = -24 m$$ $m^2$ terms: $$36 m^2 - 20 m^2 = 16 m^2$$ $m^3$ terms: $$-8 m^3 + 8 m^3 = 0$$ $m^4$ terms: $$4 m^4 - 4 m^4 = 0$$ So: $$D = 48 - 24 m + 16 m^2$$ --- 6. **Set $D=0$ for tangency:** $$48 - 24 m + 16 m^2 = 0$$ Divide entire equation by 8: $$6 - 3 m + 2 m^2 = 0$$ Rewrite: $$2 m^2 - 3 m + 6 = 0$$ Calculate discriminant of this quadratic in $m$: $$ riangle = (-3)^2 - 4 imes 2 imes 6 = 9 - 48 = -39 < 0$$ Since discriminant is negative, no real solutions for $m$. --- 7. **Re-examine step 5 for errors:** Check the expansion of the second term in step 4: $$4(1 + m^2)(4 - 2m + m^2)$$ Calculate $(1 + m^2)(4 - 2m + m^2)$: $$= 1 imes (4 - 2m + m^2) + m^2 imes (4 - 2m + m^2) = 4 - 2m + m^2 + 4 m^2 - 2 m^3 + m^4 = 4 - 2 m + 5 m^2 - 2 m^3 + m^4$$ Multiply by 4: $$16 - 8 m + 20 m^2 - 8 m^3 + 4 m^4$$ This is correct. Check the first term expansion again: $$( -8 + 2m - 2m^2 )^2 = (-8)^2 + 2 imes (-8)(2m - 2m^2) + (2m - 2m^2)^2$$ Calculate: $$64 + 2 imes (-8)(2m - 2m^2) + (2m - 2m^2)^2 = 64 - 32 m + 32 m^2 + 4 m^2 - 8 m^3 + 4 m^4 = 64 - 32 m + 36 m^2 - 8 m^3 + 4 m^4$$ This is correct. So the discriminant $D$ is: $$D = (64 - 32 m + 36 m^2 - 8 m^3 + 4 m^4) - (16 - 8 m + 20 m^2 - 8 m^3 + 4 m^4) = 48 - 24 m + 16 m^2$$ --- 8. **Solve $48 - 24 m + 16 m^2 = 0$** Divide by 8: $$6 - 3 m + 2 m^2 = 0$$ Rewrite: $$2 m^2 - 3 m + 6 = 0$$ Discriminant: $$ riangle = (-3)^2 - 4 imes 2 imes 6 = 9 - 48 = -39 < 0$$ No real roots for $m$ means no real tangent lines with slope $m$ passing through $(1, -1)$. --- 9. **Check if the line passes through $(1, -1)$ and is tangent to the circle:** Since no real $m$ satisfies the tangency condition, the line $y = mx + c$ passing through $(1, -1)$ cannot be tangent to the circle. --- 10. **Alternative approach: Use general tangent condition** The distance from the center $(4, -2)$ to the line $y = mx + c$ must equal the radius $ ho = \\sqrt{13}$. Distance formula: $$d = \frac{|m x_0 - y_0 + c|}{\sqrt{m^2 + 1}}$$ Center: $(x_0, y_0) = (4, -2)$ Substitute $c = -1 - m$: $$d = \frac{|m \times 4 - (-2) + (-1 - m)|}{\sqrt{m^2 + 1}} = \frac{|4 m + 2 - 1 - m|}{\sqrt{m^2 + 1}} = \frac{|3 m + 1|}{\sqrt{m^2 + 1}}$$ Set $d = \sqrt{13}$: $$\frac{|3 m + 1|}{\sqrt{m^2 + 1}} = \sqrt{13}$$ Square both sides: $$\frac{(3 m + 1)^2}{m^2 + 1} = 13$$ Multiply both sides by $m^2 + 1$: $$(3 m + 1)^2 = 13 (m^2 + 1)$$ Expand left: $$9 m^2 + 6 m + 1 = 13 m^2 + 13$$ Bring all terms to one side: $$9 m^2 + 6 m + 1 - 13 m^2 - 13 = 0$$ Simplify: $$-4 m^2 + 6 m - 12 = 0$$ Divide by -2: $$2 m^2 - 3 m + 6 = 0$$ This is the same quadratic as before with discriminant $-39 < 0$. No real solutions for $m$. --- **Final conclusion:** There are no real values of $m$ such that the line $y = mx + c$ passing through $(1, -1)$ is tangent to the circle $C$. --- **Answer:** (a) $$c = -1 - m$$ (b) No real values of $m$ satisfy the tangency condition, so no tangent line $L$ passes through $(1, -1)$.