Circles practice · Mixed difficulty

The correct answer is B (${x^2} + {y^2} - 2hx - 2ky + {h^2} = 0$) because it represents the equation of a circle with center (h, k) touching the x-axis.

Option D ($\frac{{1}}{{2}}$) is correct because it represents the distance from the origin to the center of the circle formed by the locus of the midpoint of the chords of contact.

Option A, 7, is correct because solving the equations derived from the tangent and normal conditions gives $\alpha = 1, \beta = -1, r = 5$, and thus $\alpha - \beta + r = 7$.

Option A (${x^2} + {y^2} - 8x - 8y + 24 = 0$) is correct because the center of the circle lies on the line $x - y = 0$, touches the line $x + y = 4$, and the coordinates of the center satisfy $x + y > 4$.

The correct answer is C. The locus of the intersection point is an ellipse, and its eccentricity is calculated to be $\frac{{\sqrt 3 }}{{2}}$.

The correct answer is B (2) because by substituting the points into the general equation of a circle, we find that k = 2 is the only valid solution.

Option C ((3,-1)) is correct because the tangent to the first circle at (1,-2) is also tangent to the second circle, and (3,-1) satisfies both the tangent line equation and the second circle's equation after shifting.

The correct answer is $\frac{{r^4}}{{2|a b|}}$ because the area of the triangle is calculated using the intercepts of the tangent line on the coordinate axes.

Option C ((0,2q)) is correct because the set of possible values of p is (0, 2q), derived from the condition that the given line is tangent to the circle.

Option A (Circle) is correct because the locus equation simplifies to the standard form of a circle, where the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term.

Option D ($\phi$) is correct because there is no value of $\lambda$ that satisfies both the condition for P to be an interior point and the condition that the circle neither touches nor intersects the axes.

Option A is correct because the radius of the circle is the square root of the constant term in the equation, which is $\sqrt{36} = 6$.

The correct answer is D ($2x - 3y = - 13$) because it represents the equation of the chord bisected at the given point.

Option C ($x^2+y^2-2 x+2 y=47$) is correct because the center of the circle is found by solving the equations of the diameters, and the radius is calculated from the area. Substituting these values into the circle equation yields the correct result.

Option B ($y = 3,12x + 5y = 39$) is correct because these lines satisfy the condition of intercepting chords of length 8 units from the given circle when passing through (2,3).

Option B (Are perpendicular) is correct because the line connecting the origin to the intersection points are perpendicular.

Option C is correct because it involves finding the equation of a circle passing through three points and then using the fourth point to solve for k.

The correct answer is $11 x-2 y-46=0$ because it is derived using the tangent equation formula $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.

Option C ($\frac{{\sqrt 7 }}{3}$) is correct because after substituting $\lambda = 3$ into the given equation and simplifying, the radius of the resulting circle is calculated to be $\frac{{\sqrt 7 }}{3}$.

Option D ($P$ lies inside $C$ but outside $E$) is correct because substituting the coordinates of P into the equations of the ellipse and circle shows that P lies inside the circle and outside the ellipse.

Option A (4) is correct because the length of the chord is calculated to be 4 using trigonometric identities and the distance formula.

Option C (Intersect at two points) is correct because the distance between the centers of the circles is less than the sum of their radii, indicating they intersect.

The correct answer is A (${x^2} + {y^2} - 6x + 6y + 9 = 0$) because it represents a circle with radius 3 in the fourth quadrant touching both axes.

Option D ($1 / \sqrt{2}$ units) is correct because the radius is calculated using the given conditions and the equation of the circle.

Option D (All of these) is correct because the line always intersects the circle, meaning the discriminant of the quadratic equation formed by substituting the line equation into the circle equation is always non-negative for all values of m.

Option A ($4 y+3 x+22=0$) is correct because it represents the equation of the incident ray, considering the reflection from the line $x=-2$ and the tangency to the circle $x^2+y^2=4$.

Option C ((5,2)) is correct because the center of the circle must be equidistant from both lines, and solving the resulting equations gives the center's coordinates.

Option C (${\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$) is correct because deriving the equation of the tangent and using the perpendicularity condition leads to this locus.

Option A (${x^2} + {y^2} - 8x = 0$) is correct because substituting the point (5, -7) into the equation results in a positive value, indicating the point lies outside the circle.

Option D ($5{x^2} + 5{y^2} + 60x + 7 = 0$) is correct because it represents the locus of a point where the ratio of tangent lengths to the two circles is 2:3.