Circles practice · Easy

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 ($k \geq \frac{1}{2}$) 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 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 (${\cos ^2}\mathrm{\alpha } = 1$) 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.

Option C ($x-y=0$) is correct because the center of the circle lies on the line equidistant from the two parallel tangent lines.

Option B ($\frac{{192}}{{25}}\;\mathrm{sq}.\mathrm{units}$) is correct because it is calculated using the formula for the area of the triangle formed by the tangents from a point to a circle and the chord of contact.

Option B ($x^2+y^2-2 x+4 y-11=0$) is correct because it represents the equation of a circle with center (1, -2) and radius 4.

Option C ($x^2+y^2-2 x+2 y-47=0$) is correct because the center of the circle is found by solving the diameters' equations, and the radius is derived from the given area.

Option D ($\sqrt{c_1-c}$) is correct because it is the length of the tangent calculated using the formula $\sqrt{S_1}$.

Option A (12) is correct because by comparing the equation of the circle with the given information, we find a = -2, b = 1, and c = -13, leading to a+b-c = 12.

The correct answer is Collinear because the determinant formed by the coordinates of the centers of the three circles is zero, indicating that they lie on the same line.

Option B ((-7,8)) is correct because the locus of P is a circle with center (-7,8) after applying the given ratio condition.

Option A (2 < r < 8) is correct because it satisfies the condition for two circles to intersect at two distinct points, where the distance between the centers is between the difference and the sum of the radii.

Option B ($\frac{a-b}{h}=\frac{a^{\prime}-b^{\prime}}{h^{\prime}}$) is correct because it represents the condition for two curves to intersect in four concyclic points.

Option A (1) is correct because the product of the roots of the equation formed by the concyclic points is 1.

Option C ($x^2+y^2-2 x-6 y=0$) is correct because it represents a circle passing through the origin with its center on the line y=3x at a distance of $\sqrt{10}$ from the origin.

Option A ($x^2+y^2+x+2 y-5=0$) is correct because it represents a circle touching the given line at (1,1) and satisfies the condition S(2,3)=16.

Option C ($x^2+y^2-4 x+2 y+1=0$) is correct because the director circle of a circle has the same center but a radius that is $\sqrt{2}$ times the original circle's radius.

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

Option A ($\sqrt{{\beta-\alpha}}$) is correct because it represents the length of the tangent calculated using the given conditions.