Circles practice · Medium

Option C (${x^2} + {y^2} - 4x - 2y + 4 = 0$) is correct because it represents the mirror image of the given circle in the line x + y = 2, found by finding the image of the center.

Option D (a circle) is correct because the locus of the point, satisfying the given condition, results in the equation of a circle.

Option D $\left( {9,\frac{{92}}{9}} \right]$ is correct because it represents the set of k values for which the given circle lies inside the fourth quadrant and the point (1, -1/3) lies on or inside the circle.

Option A is correct because the common chord equation is derived from subtracting the two circle equations, and the resulting line passes through the point (5/12, 5/12).

The correct answer is D (none of these) because the given conditions do not lead to any of the provided options for the range of 'a'.

Option A ($x^2+y^2-30 x+12 y-27=0$) is correct because it represents the equation of the circle with the normal chord PQ as diameter, calculated using the properties of parabolas and normals.

Option B ($\left(\frac{1}{2},-\sqrt{2}\right)$) is correct because it represents the center of the circle passing through the given points and touching the specified circle.

Option A ($\sqrt{5} y-2 x-4=0$) is correct because it represents the common tangent to both the circle and the hyperbola, satisfying the tangency conditions for both curves.

Option B ($120(x^2+y^2)-22(15 y+8 x)+289=0$) is correct because the equation of the circle with the given diameter is found by using the roots of the two quadratic equations.

Option D (${x^2} + {y^2} + x + y = 0$) is correct because it represents the circle described on the common chord of the given circles as a diameter, which is found by using the equation $S_1 + \lambda(S_1 - S_2) = 0$ and the condition that the center lies on the common chord.

Option B ($x^2+y^2-3 x-5 y+8=0$) is correct because it represents the circumcircle of the triangle formed by the given lines, found by using the general equation of a circle passing through the vertices of the triangle.

Option C (72) is correct because by substituting the point and using the tangency condition, we can derive the value of 'c' as 72.

Option B ($x + y = 2\sqrt 2$) is correct because it satisfies the condition of being tangent to the given circle and having equal intercepts on the positive coordinate axes.

Option D ($(-\sqrt{{2}},-1) \cup (1, \sqrt{{2}})$) is correct because the condition $\frac{{\pi}}{{2}} < \theta < \pi$ implies that $\tan \theta < 0$, which leads to the given range for 'a'.

Option B (8) is correct because solving the equation derived from the condition that the circles touch each other yields k = 8.

Option B ($x^2+y^2-4 x-10 y+19=0$) is correct because CP is the diameter of the circumcircle of triangle PAB, and the equation is derived accordingly.

The correct answer is B $\left(\frac{{2 a^2 b}}{{{{a^2+b^2}}}}, \frac{{2 a b^2}}{{{{a^2+b^2}}}}\right)$ because it represents the reflection of the origin in the line AB, calculated using perpendicularity and equidistance conditions.

Option B ${x^2} + {y^2} + 6x + 2y - 15 = 0$ is correct because it represents a circle with radius 5, touching the given line at (1,2), and having a portion in all quadrants.

Option A (circle of radius 2) is correct because the locus of P is a circle with radius 2, derived from the perpendicular tangent condition.

The correct answer is D (None of these) because the perpendicular distance from the center of the circle to the line is greater than the radius, meaning the line lies outside the circle.

Option B ($\sum {y_i} = - 4\left( {f + 2a} \right)$) is correct because it is derived from Vieta's formulas applied to the quartic equation obtained by substituting the parabola equation into the circle equation.

Option D ($\frac{\pi}{2}$) is correct because point P lies on the director circle of the given circle, implying tangents from P are perpendicular.

Option C $\left(\frac{1}{4}, \frac{1}{2}\right)$ is correct because it involves finding the point on the circle closest to the parabola, which requires analyzing the normal to the parabola and its intersection with the circle.

Option A ($\left(-\alpha, \frac{{1}}{{\beta}}\left(\alpha^2-3\right)\right)$) is correct because the polar equation must be independent of k, leading to two equations that can be solved for the fixed point.

Option C ((3,3)) is correct because the chord of contact for tangents drawn from a point on the line x+y=3 to the circle x²⁺y²=9 always passes through the point (3,3).

Option A (2) is correct because the radius of the second circle can be found using the condition for touching circles and then multiplied by 4.

Option B (They intersect each other) is correct because the distance between the centers is less than the sum of the radii and greater than the absolute difference of the radii.

Option C ($x-3 y+15=0$) is correct because the equation of the chord of contact is found by using the formula T=0, where T is obtained by substituting the point (1,8) into the circle equation.

Option D ($6 \sqrt{2}+6$) is correct because it correctly calculates the radius of the circle tangent to the line pair and the line y=6.

Option A ($x^{2}+y^{2}-2 x-2 y-14=0$) is correct because it represents the circumcircle equation derived from the incenter and side equation of the equilateral triangle.