Question

Consider the circle $x^2 + y^2 - 8x + 2y - 8 = 0$. Let C be the center of the circle. Also, the tangents to this circle at A(1, -5) and B(8, -4) meet at P. Let S = O represent the family of circles passing through A and B, then

• A. area of quadrilateral CAPB is 25 sq. units
• B. the smallest possible circle of the family S = O is $x^2 + y^2 - 9x + 9y + 28 = 0$
• C. the radical axis of the family of circles S = O is $x - 7y - 36 = 0$
• D. sum of the squares of the abscissa and ordinate of point P is 89

Answer 1

Answer: (A), (B), (C), (D)

All four options are correct.

Answer 2

The given circle is $x^2 + y^2 - 8x + 2y - 8 = 0$.

The center of the circle C is found by comparing the equation with $x^2 + y^2 + 2gx + 2fy + c = 0$, where $2g = -8 \implies g = -4$ and $2f = 2 \implies f = 1$. The center is $(-g, -f) = (4, -1)$.

The radius squared is $r^2 = g^2 + f^2 - c = (-4)^2 + 1^2 - (-8) = 16 + 1 + 8 = 25$. The radius is $r = 5$.

So, C(4, -1) and radius = 5.

The points A(1, -5) and B(8, -4) are on the circle.

The tangents to the circle at A and B meet at P. P is the pole of the chord AB.

The equation of the tangent at A(1, -5) is $x(1) + y(-5) - 4(x+1) + 1(y-5) - 8 = 0$, which simplifies to $x - 5y - 4x - 4 + y - 5 - 8 = 0 \implies -3x - 4y - 17 = 0 \implies 3x + 4y + 17 = 0$.

The equation of the tangent at B(8, -4) is $x(8) + y(-4) - 4(x+8) + 1(y-4) - 8 = 0$, which simplifies to $8x - 4y - 4x - 32 + y - 4 - 8 = 0 \implies 4x - 3y - 44 = 0$.

To find P, we solve the system of equations:

1. $3x + 4y = -17$

2. $4x - 3y = 44$

Multiplying (1) by 3 and (2) by 4: $9x + 12y = -51$ and $16x - 12y = 176$.

Adding these equations: $25x = 125 \implies x = 5$.

Substituting $x=5$ into (1): $3(5) + 4y = -17 \implies 15 + 4y = -17 \implies 4y = -32 \implies y = -8$.

So, P is (5, -8).

Area of quadrilateral CAPB: C(4, -1), A(1, -5), P(5, -8), B(8, -4). CA and CB are radii (length 5). PA and PB are tangents from P to the circle. CA is perpendicular to PA, and CB is perpendicular to PB. Thus, $\angle CAP = \angle CBP = 90^\circ$. The quadrilateral CAPB is a kite. The area of the kite is the sum of the areas of the two congruent right-angled triangles $\triangle CAP$ and $\triangle CBP$.

$CA = 5$. $PA = \sqrt{(5-1)^2 + (-8-(-5))^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$.

Area of $\triangle CAP = \frac{1}{2} \times CA \times PA = \frac{1}{2} \times 5 \times 5 = \frac{25}{2}$.

Area of CAPB = $2 \times$ Area of $\triangle CAP = 2 \times \frac{25}{2} = 25$ sq. units.

The smallest possible circle of the family S = O (circles passing through A and B) is the circle with AB as diameter.

The center of this circle is the midpoint of AB. Midpoint M of A(1, -5) and B(8, -4) is $(\frac{1+8}{2}, \frac{-5-4}{2}) = (\frac{9}{2}, -\frac{9}{2})$.

The radius squared of this circle is $(\frac{\text{length of AB}}{2})^2 = \frac{(8-1)^2 + (-4-(-5))^2}{4} = \frac{7^2 + 1^2}{4} = \frac{49+1}{4} = \frac{50}{4} = \frac{25}{2}$.

The equation of the circle is $(x - \frac{9}{2})^2 + (y - (-\frac{9}{2}))^2 = \frac{25}{2}$.

$(x - \frac{9}{2})^2 + (y + \frac{9}{2})^2 = \frac{25}{2}$

$x^2 - 9x + \frac{81}{4} + y^2 + 9y + \frac{81}{4} = \frac{25}{2}$

$x^2 + y^2 - 9x + 9y + \frac{162}{4} = \frac{25}{2}$

$x^2 + y^2 - 9x + 9y + \frac{81}{2} = \frac{25}{2}$

$x^2 + y^2 - 9x + 9y + \frac{81-25}{2} = 0$

$x^2 + y^2 - 9x + 9y + \frac{56}{2} = 0$

$x^2 + y^2 - 9x + 9y + 28 = 0$.

The radical axis of the family of circles S = O (passing through A and B) is the line passing through A and B.

The equation of the line passing through A(1, -5) and B(8, -4) is:

$\frac{y - (-5)}{x - 1} = \frac{-4 - (-5)}{8 - 1}$

$\frac{y + 5}{x - 1} = \frac{1}{7}$

$7(y + 5) = 1(x - 1)$

$7y + 35 = x - 1$

$x - 7y - 36 = 0$.

Sum of the squares of the abscissa and ordinate of point P.

P is (5, -8).

Sum of squares of coordinates = $5^2 + (-8)^2 = 25 + 64 = 89$.