Circles practice · Mixed difficulty

Option C (3) is correct because the distance between the centers of the circles equals the sum of their radii, indicating that the circles touch externally, hence having 3 common tangents.

Option B ((3,-1)) is correct because the tangent to the first circle at (1,-2) is also tangent to the second circle at (3,-1).

Option D (None of these) is correct because the range of $\lambda$ for which the equation represents a circle is $[-2, -1]$.

Option A (cuts the given circle orthogonally) is correct because the condition for the chord of contact passing through Q leads to the orthogonality condition of the two circles.

Option A ((10,0)) is correct because it minimizes the area of the triangle formed by the tangents to the circle from a point on the x-axis and their y-intercepts.

Option C ($x{x_1} + y{y_1} = x_1^2 + y_1^2$) is correct because it represents the equation of the chord of the circle with the given midpoint.

Option C ((2,2)) is correct because converting the polar equation to Cartesian coordinates and completing the square reveals the center of the circle.

Option C ($\frac{12}{5}$) is correct because it is the value of $r^2$ that satisfies the given condition: the chord length is equal to r.

Option B Touch each other internally is correct because the distance between the centers is equal to the difference of the radii of the two circles.

Option A ($ab > 0, c > 0$) is correct because for one circle to lie within the other, their intersection points must be imaginary and the origin must be outside both circles.

Option A ($h = \pm r$) is correct because the condition for perpendicular tangents from the origin leads to this relationship between h and r.

Option C ($- 1$) is correct because the slope of the tangent to the circle at point (h,h) is -1.

Option B (-3,-3) is correct because the point of contact is found by solving the equation of the normal line passing through the circle's center and the given line.

Option C $\left( {5,2} \right)$ is correct because the symmetrical point about a line is the image of the point in that line, and the tangent at the origin is $y=x$.

Option D ($24 x+7 y=0$) is correct because it represents the equation of the other equal side passing through the origin, found using the perpendicular distance from the center to the line.

Option A ($4\sqrt 2 - 5$) is correct because it represents the shortest distance between the parabola and the circle, calculated along the common normal.

Option B ($l_1^2+l_2^2$) is correct because when points are conjugate with respect to the circle, the square of the distance between them equals the sum of the squares of the lengths of the tangents from the points to the circle.

Option C $((-3 \sqrt{2},-\sqrt{3}) \cup(\sqrt{3}, 3 \sqrt{2}))$ is correct because it represents the set of 'a' values that satisfy both conditions: the point lies outside the first circle and inside the second circle.

Option A ($(1 / 2,1 / 4)$) is correct because the chord of contact equation can be rearranged to represent a family of lines passing through the intersection of two fixed lines, which gives the required point.

The correct answer is $x+y=2 \sqrt{{2}}$ because the perpendicular distance from the center of the circle to the tangent must equal the radius.

Option D ($\pi / 2$) is correct because the origin lies on the director circle of the given circle.

Option C ($m \in(-2,-1 / 2)$) is correct because it represents the range of slopes for which the line does not intersect the circle, determined by ensuring the distance from the circle's center to the line is greater than the radius.

Option A ($2 g^{\prime}\left(g-g^{\prime}\right)+2 f^{\prime}\left(f-f^{\prime}\right)=c-c^{\prime}$) is correct because it represents the condition for the first circle to bisect the circumference of the second circle.

Option C (16) is correct because there are 16 integer values of 'a' that satisfy the condition that the radius of the circle does not exceed 5.

Option B ($x - y = 0$) is correct because it satisfies the condition that the intercepts made by the circle on L1 and L2 are equal.

Option B (7) is correct because the radius is found by converting the polar equation to Cartesian form and then calculating the radius of the resulting circle.

Option A ($4{x^2} + 4{y^2} - 4x - 4y + 1 = 0$) is correct because it represents a circle that touches both axes and the given line in the first quadrant, lying below the line.

Option A ($-1 < m < \frac{3}{4}$) is correct because the origin divides AB internally when the line y=mx lies between the tangents at the origin to the two circles.

Option D (None of these) is incorrect, and Option A ($y = m(x - 1) - 2 \pm 3\sqrt{1 + m^2}$) is the correct equation of the tangent to the circle, which is not listed as an option. The provided answer is missing a minus sign and has an extra square.

Option C ($\left(\frac{{1}}{{2}}, \frac{{1}}{{4}}\right)$) is correct because the chord of contact passes through this fixed point, determined by substituting the point P on the line into the chord of contact equation.