Circles practice · Mixed difficulty

Option C ($\sqrt {{{a^2}(1 + {m^2})}} > c$) is correct because the condition for the line to intersect the circle in real points is that the discriminant of the quadratic equation formed by substituting the line equation into the circle equation must be greater than or equal to zero.

Option C (No value of $a$) is correct because the condition for the line to pass through the intersection points leads to a quadratic equation with no real solutions for $a$.

Option B (${x^2} + {y^2} - 30x - 10y + 225 = 0$) is correct because it represents a circle with radius 5, centered in the first quadrant, touching the x-axis and the line 4y = 3x.

Option C ($9(x^2+y^2)=a^2$) is correct because the straight line touches the circle with radius a/3, derived from the condition that the pole lies on the circle $x^2+y^2=9a^2$.

Option C ((-4,0)) is correct because the range of g is determined by the condition that the point (α, α) lies outside the circle for all real α.

Option C (19) is correct because the point C(a, a+1) must lie outside the circle x² + y² = 4, leading to 19 integral values for 'a' within the given range.

Option A ($2 x^2+2 y^2-a x-a y=0$) is correct because it represents the circle with the common chord of the given circles as its diameter.

Option C ($\frac{10}{9} \sqrt{5}$ sq units) is correct because it is calculated using the length of the perpendicular from the center to the chord and the length of the chord.

Option B (-11) is correct because substituting the point (-3, 2) into the concentric circle equation allows us to solve for c.

Option A ((2,-3), 1) is correct because by completing the square, the center and radius of the circle $x^2+y^2-4 x+6 y+12=0$ are found to be (2, -3) and 1, respectively.

Option A ($\pm 5$) is correct because the condition for tangency (perpendicular distance from center to line equals radius) leads to $c = \pm 5$.

Option C (a circle of radius $\sqrt{a^2-r^2}$) is correct because the locus of point P satisfies the equation of a circle centered at the origin.

Option A ($x^2+y^2-8 x-6 y+16=0$) is correct because it satisfies the conditions: radius 3, center on the line $y=x-1$, and passing through the point (7,3).

Option C (${x^2} + {y^2} - 3\;x - 4\;y = 0$) is correct because it represents the equation of the circle passing through the origin and making the specified intercepts.

Option D ($x^2+y^2-2 x-2 y+1=0$) is correct because it satisfies the given conditions for S(x,2)=0 and S(1,y)=0.

Option C ($p^2-q^2=0$) is correct because the tangents from the origin are perpendicular only when the circle touches both x and y axes, implying $p = \pm q$.

Option D ($k \geq \frac{1}{2}$) is correct because the condition for the circle to be tangent to the x-axis and pass through (-1,1) leads to a quadratic equation with a real solution for the center's x-coordinate, implying the discriminant must be non-negative.

Option A ($x^2+y^2-6 x \pm 8 y+9=0$) is correct because it represents the equations of the circles that touch the x-axis at a distance of 3 from the origin and have a y-intercept of length $2 \sqrt{7}$.

Option A ($(8, \infty)$) is correct because the condition for the line to bisect two distinct chords leads to the inequality $a^2 > 8$.

The correct answer is B (${x^2} + {y^2} - ax = 0$) because it represents the locus of the center of the circle drawn on the chord as diameter.

Option C (a hyperbola) is correct because the difference in distances from the center of the moving circle to the centers of the fixed circles is constant, which is the defining property of a hyperbola.

Option A (0) is correct because the point (2,2) lies inside the circle, so no real tangents can be drawn.

Option A (3) is correct because the distance between the centers is equal to the sum of the radii, indicating that the circles touch externally, hence 3 common tangents.

Option A ($x+2 y=0$) is correct because it is the locus of the centers of the circles passing through the intersection of the given circles.

Option B (G.P.) is correct because the lengths of the tangents are in G.P. as shown by the relationship derived from the given conditions.

Option B ($6 + 4\sqrt 2$) is correct because it is the solution to the quadratic equation derived from the geometry of the two touching circles.

Option B (${\cos ^2}\mathrm{\alpha } = 1$) is correct because it represents the condition for tangency when the given line is tangent to the circle, derived using the formula c² = a²(1 + m²).

Option A (1) is correct because it is the only value of $\lambda$ that satisfies both conditions for the equation to represent a circle.

Option D (25,29) is correct because the point (1,4) must satisfy the inequality for lying inside the circle, and the radius must be less than the distance to the axes.

The correct answer is D. The line being normal to the circle implies it passes through the center. Only option D has a center that satisfies the normal's equation and the given point lying on the circle.