# Circle Problems

### Circle Problems

1) Show that the line x + y + 1 = 0 touches the circle x² + y² – 3x + 7y + 14 = 0 and find the point of contact.

Solution: Centre of the circle is C (3/2, -7/8)

Radius of the circle (r) = $$\sqrt{\frac{9}{4}\,+\,\frac{49}{4}\,-\,14}\,=\,\sqrt{\frac{58\,-56}{4}}\,=\,\sqrt{\frac{1}{2}}\,=\frac{1}{\sqrt{2}}$$

Perpendicular distance from the centre to the given line = $$\frac{\left| \frac{3}{2}\,-\,\frac{7}{2}\,+1 \right|}{\sqrt{1\,+\,1}}\,=\,\frac{1}{\sqrt{2}}\,=\,r$$

∴ The line x + y + 1 = 0 touches the given circle.

If P (h, k) is the point of contact then $$\frac{h\,-\,\frac{3}{2}}{1}\,=\,\frac{k\,+\,\frac{7}{2}}{1}\,=\,\frac{-\left( \frac{3}{2}\,-\,\frac{7}{2}\,+\,1 \right)}{1\,+\,1}$$

⇒ h – 3/2 = k + 7/2 = ½

⇒ h – 3/2 = ½, k + 7/2 = ½

⇒ h = 2, k = -3

∴ Point of Contact P (2, -3)

2) Find the equation of the circle passing through (0, 0) and making intercept 6 units on x – axis and intercept of 4 units on y-axis.

Solution: Let the circle cuts x – axis at O, A and y – axis at O, B Such that OA = 6, OB = 4.

Let C be the center of the circle and D be the midpoint of OA. Then OD = 3, CD = OB/2 =2.

Now $$OC\,=\,\sqrt{O{{D}^{2}}\,+\,C{{D}^{2}}}\,=\,\sqrt{9\,+\,4}$$. Radius of the circle = √13

Centre of the circle is C (±3, ±2)

Required circle equation is (x ± 3)² + (y ± 2)² = (√13)²

⇒ x² + y² ± 6x ± 4y = 0