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Since P(8, k) is the midpoint of the chord, the chord is perpendicular to OP. Slope of OP = k/8. Therefore, the slope of the chord is -8/k which is an integer if  k = ±1,±2,±4 and ±8. If k = ±8, then the point P lies outside the circle so that it cannot be the midpoint of any chord. Therefore

k ≠ ±8

k = ±1,±2,±4

so that the integer values of k is 6.

Correct option  (A)  14/5

Explanation :

It is known that point P is the internal centre of similitude and AP:PB = 2:3 . Therefore,

3AP = 2PB = 2(AB - AP) = 2(7 - AP)

⇒ 5 AP = 14

⇒ AP = 14/5

Let

S' ≡ x² + y² - 9 = 0

S'' ≡ x² + y² - 2x + 8y - 7 = 0

Let S ≡ x² + y² + 2gx + 2fy + c = 0 be the required circle. Since S = 0 passes through (1, 2), we have

2g + 3f + c = -5  ....(1)

S = 0 bisects the circumference of S' = 0

S − S' = 0 passes through (0, 0). This gives

c + 9 = 0

c = -9  .....(1)

S = 0 and S'' = 0 cut orthogonally which implies that

2g(-1) + 2f(4) = c - 7

2g + 8f = c - 7   ....(2)

From Eqs. (1)–(2), we have

herefore, g = 4,f = -1, and . c =9 Hence the required circle is

Let S ≡ x² + y² + 8x - 2y - 9 = 0

Let S ≡ x² + y² + 2gx + 2fy + c = 0 be the required circle. Since it passes through the origin (0, 0), c = 0. Put y = x in S = 0. Then x² + (g + f )x = 0. Therefore, x = 0, x = −(g + f ) so that the points of intersection are A(0, 0) and B[−(f + g), −(f + g)]. Now

AB = √2

f + g = ±1

Similarly

g - f  = ±

Therefore, the centres of circles are given by (1, 0), (−1, 0), (0, 1) and (0, −1) and the equations of the circle are given by + x² + y² ± 2x = 0 and x² + y² ± 2y = 0.

If k = 1, then PA = PB so that the locus of P is the perpendicular bisector of (bar)AB.

Suppose, there are three rational points on circle C. Now consider that the triangle with those rational point vertices whose circumcenter is (0, √2). Since the vertices of this triangle are rational points, its equations of the perpendicular bisectors are first-degree equations in x and y with rational coefficients and hence the circumcentre must be a rational point, but here it is not a rational point because (0, √2) is the circumcentre. Hence, maximum number of rational points on the circle is 2.

Equation of the circle passing through points P, Q, R and S is of the form

(λx - y + 1)(x - 2y + 3) + μ(xy) = 0

This equation represents a circle if the coefficient of x² is equal to the coefficient of y² and coefficient of xy = 0. Therefore,

λ = 2 and μ = 5

Hence, the value of λ = 2

Let S ≡ x² + y² − 2ax = 0 and P = (h, k) be a point on S = 0. Therefore

h² + k² - 2ah = 0

Now, Q (h/2 ,k/2 ) is the centre of the circle drawn on OP as the diameter. we have

(h/2)² + (k/2)² - a(h/2) = 0

 Therefore, the locus of Q is x² + y² − ax = 0. 

Let S ≡ x² + y² + 2gx + 2fy + c = 0 be the required circle. It passes through (0, 0). This implies c = 0. Now S = 0 touches the line x -  y = 0. Therefore

If S = 0 bisects the circumference of S'  ≡ x² + y² + 2y − 3 = 0, then S - S' = 0 passes through the centre (0, 1) of S' = 0. This implies that

2gx - 2(g + 1)y + 3 = 0

passes through (0, -1 )   (:. f = -g)

⇒ 0 -2(g + 1)(-1) + 3 = 0

⇒2g = -5 = -2f

Therefore, S ≡ x² + y²  - 5x + 5y = 0

Let P(x₁, y₁) be a point on x² + y² = R². Suppose the tangents from point P to the circle x² + y² = r² meet the circle with radius R in A and B such that AB touches the circle with centre r. Thus, for ΔPAB, x² + y² = R² is the circumcircle and x² + y² = r² is the incircle and hence the circumcentre and incentre are the same. Therefore, the triangle is equilateral so that R= 2r

 We have

S ≡ x² + y² - a² = 0

and (x₁,y₁) =(a cosθ, a sin θ)

 The equation of the tangent at (a cos θ, a sin θ) is

x(a cosθ) + y(a sinθ) - a² = 0

x cosθ + y sinθ - a = 0

 We have

2² + 3² - 2(2) - 4(3) + = 16 - 16 = 0

 which implies that (2, 3) lies on the circle S = x² + y² − 2x − 4y +3 = 0. Here, (x₁, y₁) = (2, 3) so that the equation of the tangent at (2, 3) is

 S₁ = x(2) + y(3) − (x + 2) − 2(y + 3) + 3 = 0 that is, S₁ = x + y − 5 = 0.

 We know that y-axis touches the circle x² + y² + 2gx + 2fy + c = 0 if f² = c. For the circle mentioned in this problem also, we have q² = f² = c = q². Therefore, y-axis touches the circle. Since the tangents drawn from the origin are at right angles, the other tangent touches the x-axis. Therefore,

 p² = q²

 Aliter: Since the tangents drawn from (0, 0) are at right angles, origin must lie on the director circle of the given circle Therefore, (0, 0) lies on the circle (x − p)² + (y − q)² + 2p².

Correct option (a,d)

Explanation :

If the tangents drawn are at right angles, the points must be the intersection of the line x = 2 with the director circle x² + y² = 32 of the circle x² + y² = 16. Substituting x + 2 in x² + y² = 32, we have

y² = 28 or y = ±2√7

Therefore, the points are (2, 2√7 ) and (2, -2√7).

Let S ≡ x² + y² + 2gx + 2fy + c = 0 be the circle passing through points A, B and C. Therefore

2f + c = -1

4g + 6f + c = -13

-4g + 10f + c = -13

-4g +10f + c = -29     .......(1)

Solving the system of equations provided in Eq. (1), we get get g = 1/3, f = −10/3 and c = 17/3 so that the equation of the circle is

x¹ + y² + 2/3x - 20/3y + 17/3 = 0

3x² + 3y² + 2x - 20y + 17 = 0

Note: Under the given hypothesis, to find the equation of the circle, it is sufficient if we find its centre and radius or assume the circle as x² + y² + 2gx + 2fy + c = 0 and find the values of g, f and c.

P(h, k) is the centre of the circle passing through points A, B, C and D (Fig.) where AB = 2a and DC = 2b. This implies and is implied by

 PA - PC -  radius of the circle

(PA)² = (PC)²

(AM)² + (PM)² = (PA)² = (PC)² = (PN)² + (CN)²

 where M and N are the midpoints of AB and CD, respectively. From the above, we have

a² + k² = h² + b²

k² - h² = b² - a²

Locus of (h,k) is y² - x² = b² - a²

Correct option (b,d)

Explanation :

Since (3, 4) lies outside the circle x² + y² = 1, one circle has the external contact with x² + y² = 1 and the other circle has the internal contact. O = (0, 0) and r₁ = 1. Let A = (4, 3) and r₂ be the radius of the required circle OA = 5.

Case 1: r₂ = 5 -   1 = 4 (i.e. external contact). Hence, the required circle is

(x - 4)² + (y - 3)² = 16

⇒ x² + y² - 8x -6y + 9 = 0

Case 2:  r₂ = 5 + 1 = 6 (i.e. internal contact). Hence, the required circle is

(x - 4)² + (y - 3)² = 36

⇒ x² + y² - 8x - 6y -11 = 0

Let the given circles be C₁, C₂ and C₃, respectively. Let C be the required circle and its equation be S ≡ x² + y² + 2gx + 2fy + c = 0. Let S' ≡ x² + y² +1 = 0. Since C bisects the circumference of C₁, the line S − S' ≡ 2gx + 2fy + c +1 = 0 passes through the centre of C₁ = (0, 0). Therefore,

c = 1

Now, C bisects the circumference of C₂ 

the line S − S'' ≡ 2(g − 1)x + 2fy + c +3 = 0 passes through the centre (−1, 0) of C₂. Therefore

2(g - 1)(-1) - 1 + 3 = 0

-2g + 2 - 1 + 3 = 0

g = 2

 Similarly, since the circle C bisects the circumference of circle C₃, we have f = 2. Therefore, equation of the required circle C is

S ≡ x² + y² + 4x + 4y - 1 = 0

Clearly, (a, 0) lies on the circle. Equation of the tangent at (a, 0) is 

x(a) + y(0) + a cot α (y + 0) – a² = 0 

ax + α(c cotα)y − a² = 0 

x + y cot α - a = 0

Correct option  (D)  x²  + y² - 2x - 2y = 0

Explanation :

Let S≡ x² + y² + 2gx + 2fy + c = 0 be the required circle which passes through (0, 0). This implies that

c = 0  ....(1)

The circle has the centre on the line y = x which implies that

-g = -f     ....(2)

g = f   

The circle cuts the circle x² + y² - 4x - 6y + 10 = 0 orthogonally implies that

2(g)(-2) +2f(-3) = (-3) = c + 10

-4g -6f = c + 10  .....(3)

From Eqs. (1) – (3), we have g = f = -1 and c = 0. Therefore 

S ≡ x² + y² - 2x - 2y = 0

Let S ≡ x² + y² + 2gx + 2fy + c = 0 be the required circle. S = 0. The centre (−g, −f) lies on the line x + y = 4 implies that

-g - f = 4    ....(1)

S = 0 cuts the circle S' ≡ x² + y² − 4x + 2y + 4 = 0 implies that

2g(-2) + 2f(1) = c + 4

-4g + 2f = c + 4

-4g + 2f = 4  .......(2)

Solving Eqs. (1) and (2), we have g = −2 and f = −2. Therefore

S ≡ x² + y² - 4x - 4y = 0

Correct option  (a)  4,7

Explanation :

The lines are x = 2, x = 6 and y = 5, y = 9. Therefore, the vertices of the square are (2, 5), (2, 9), (6, 9) and (6, 5).

Centre of the circle = Centre of the square

=(2 + 6/2, 9 + 5/2)

= (4,7)

The equation

x² + y² - 4x - 4y = 0   ...(i)

 represents a circle with centre (2, 2) and radius √8 = 2√2 . Then the locus of point from which perpendicular tangents can be drawn to the circle given in Eq. (1) is a concentric circle whose radius is 2 times the radius of the circle [this circle is called the director circle of the circle given in Eq. (1).

Hence, the radius of the director circle is √2(2√2) = 4. Thus k = 8.

Squaring and adding the given equations, we have x² + y² = p² + q² which represents circle with centre as origin and radius √p² + q²