Circle Tangency 1D807B

1. **State the problem:** We have two circles touching externally. Circle A has equation $x^2 + y^2 - 2x - 2y = 4$ and circle B has equation $(x - a)^2 + (y - 14)^2 = 100$. We need to find the centre and radius of circle A, then find the value(s) of $a$ such that the circles touch externally. 2. **Find centre and radius of circle A:** The general form of a circle is $x^2 + y^2 + Dx + Ey + F = 0$. Rewrite circle A's equation: $$x^2 + y^2 - 2x - 2y = 4$$ Bring all terms to one side: $$x^2 - 2x + y^2 - 2y - 4 = 0$$ Complete the square for $x$ and $y$: $$x^2 - 2x = (x - 1)^2 - 1$$ $$y^2 - 2y = (y - 1)^2 - 1$$ Substitute back: $$(x - 1)^2 - 1 + (y - 1)^2 - 1 - 4 = 0$$ Simplify: $$(x - 1)^2 + (y - 1)^2 - 6 = 0$$ So, $$(x - 1)^2 + (y - 1)^2 = 6$$ Therefore, centre of circle A is $A = (1,1)$ and radius is $r_A = \sqrt{6}$. 3. **Identify centre and radius of circle B:** Given: $$(x - a)^2 + (y - 14)^2 = 100$$ Centre of circle B is $B = (a, 14)$ and radius is $r_B = 10$. 4. **Condition for external tangency:** Two circles touch externally if the distance between their centres equals the sum of their radii: $$|AB| = r_A + r_B$$ Calculate distance between centres: $$|AB| = \sqrt{(a - 1)^2 + (14 - 1)^2} = \sqrt{(a - 1)^2 + 13^2} = \sqrt{(a - 1)^2 + 169}$$ Set equal to sum of radii: $$\sqrt{(a - 1)^2 + 169} = \sqrt{6} + 10$$ 5. **Solve for $a$:** Square both sides: $$(a - 1)^2 + 169 = (\sqrt{6} + 10)^2$$ Expand right side: $$(\sqrt{6} + 10)^2 = 6 + 2 \times 10 \times \sqrt{6} + 100 = 106 + 20\sqrt{6}$$ So: $$(a - 1)^2 + 169 = 106 + 20\sqrt{6}$$ Subtract 169: $$(a - 1)^2 = 106 + 20\sqrt{6} - 169 = -63 + 20\sqrt{6}$$ 6. **Check if right side is positive:** Calculate approximate value: $20\sqrt{6} \approx 20 \times 2.4495 = 48.99$ So: $-63 + 48.99 = -14.01$ (approximately) This is negative, which is impossible for a square. 7. **Re-examine the problem:** Since the right side is negative, no real $a$ satisfies the external tangency with the sum of radii. Check if the circles touch internally: Distance between centres equals difference of radii: $$\sqrt{(a - 1)^2 + 169} = |r_B - r_A| = |10 - \sqrt{6}|$$ Calculate right side: $$10 - \sqrt{6} \approx 10 - 2.4495 = 7.5505$$ Square both sides: $$(a - 1)^2 + 169 = (7.5505)^2 = 57.02$$ Subtract 169: $$(a - 1)^2 = 57.02 - 169 = -111.98$$ Still negative, no solution. 8. **Conclusion:** Since the problem states the circles touch externally, the only way is if the distance between centres equals sum of radii. The negative value suggests a mistake in the problem or the radius of circle A. Recalculate radius of circle A: Radius squared is 6, so radius is $\sqrt{6} \approx 2.4495$. Sum of radii: $$2.4495 + 10 = 12.4495$$ Distance between centres: $$\sqrt{(a - 1)^2 + 169} = 12.4495$$ Square both sides: $$(a - 1)^2 + 169 = 12.4495^2 = 155$$ Subtract 169: $$(a - 1)^2 = 155 - 169 = -14$$ Still negative. 9. **Check problem statement:** The problem states the circles touch externally at one point. Given the radius of circle B is 10, and circle A radius is $\sqrt{6}$, the distance between centres must be $10 + \sqrt{6}$. Since $169 = 13^2$, the vertical distance between centres is fixed at 13. Distance formula: $$\sqrt{(a - 1)^2 + 169} = 10 + \sqrt{6}$$ Square both sides: $$(a - 1)^2 + 169 = (10 + \sqrt{6})^2 = 100 + 20\sqrt{6} + 6 = 106 + 20\sqrt{6}$$ Subtract 169: $$(a - 1)^2 = 106 + 20\sqrt{6} - 169 = -63 + 20\sqrt{6}$$ Approximate: $20\sqrt{6} \approx 48.99$ So: $-63 + 48.99 = -14.01$ Negative again. 10. **Try the other possibility: the circles touch internally:** Distance between centres equals difference of radii: $$\sqrt{(a - 1)^2 + 169} = |10 - \sqrt{6}| = 10 - 2.4495 = 7.5505$$ Square both sides: $$(a - 1)^2 + 169 = 7.5505^2 = 57.02$$ Subtract 169: $$(a - 1)^2 = 57.02 - 169 = -111.98$$ Still negative. 11. **Check if the problem has a typo or if the radius of circle B is 10 or 100:** Given equation of circle B is: $$(x - a)^2 + (y - 14)^2 = 100$$ Radius squared is 100, so radius is 10. 12. **Try to find $a$ such that the circles intersect at exactly one point (tangency):** Distance between centres $d = \sqrt{(a - 1)^2 + 169}$. For external tangency: $$d = r_A + r_B = \sqrt{6} + 10$$ For internal tangency: $$d = |r_B - r_A| = 10 - \sqrt{6}$$ Since both attempts gave negative values for $(a - 1)^2$, no real $a$ satisfies these. 13. **Re-express the problem:** The problem states the circles touch externally, so the distance between centres is sum of radii. Distance squared: $$d^2 = (a - 1)^2 + 169$$ Sum of radii squared: $$(r_A + r_B)^2 = (\sqrt{6} + 10)^2 = 106 + 20\sqrt{6}$$ Set equal: $$(a - 1)^2 + 169 = 106 + 20\sqrt{6}$$ Rearranged: $$(a - 1)^2 = 106 + 20\sqrt{6} - 169 = -63 + 20\sqrt{6}$$ Calculate $20\sqrt{6} \approx 48.99$: $-63 + 48.99 = -14.01$ Negative, no real solution. 14. **Conclusion:** No real value of $a$ satisfies the external tangency condition with given data. **Final answers:** - Centre of circle A: $(1,1)$ - Radius of circle A: $\sqrt{6}$ - No real value of $a$ satisfies the external tangency condition given the problem data. If the problem expects a solution, it may require rechecking the problem data or assumptions.